Chapter 8 Resource Masters - · PDF fileThe Chapter 8 Resource Masters includes the core materials needed for Chapter ... students on how to assess performance on ... Mid-Chapter Quiz

Chapter 8 Resource Masters

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Geometry program. Any other reproduction, for sale or other use, is expressly prohibited.

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4 5 6 7 8 9 10 11 12 REL 19 18 17 16 15 14 13 12 11

CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters

booklets are available as consumable workbooks in both English and Spanish.

ISBN10 ISBN13

Study Guide and Intervention Workbook 0-07-890848-5 978-0-07-890848-4

Homework Practice Workbook 0-07-890849-3 978-0-07-890849-1

Spanish Version

Homework Practice Workbook 0-07-890853-1 978-0-07-890853-8

ANSWERS FOR WORKBOOKS The answers for Chapter 8 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

StudentWorks PlusTM This CD-ROM includes the entire Student Edition text along with the English workbooks listed above.

TeacherWorks PlusTM All of the materials found in this booklet are included for viewing, printing,

and editing in this CD-ROM.

Spanish Assessment Masters (ISBN10: 0-07-890856-6, ISBN13: 978-0-07-890856-9) These masters contain a Spanish version of Chapter 8 Test Form 2A and Form 2C.

iii

Contents

Teacher’s Guide to Using the Chapter 8

Resource Masters .........................................iv

Chapter Resources Chapter 8 Student-Built Glossary ...................... 1

Chapter 8 Anticipation Guide (English) ............. 3

Chapter 8 Anticipation Guide (Spanish) ............ 4

Lesson 8-1Geometric Mean

Study Guide and Intervention ............................ 5

Skills Practice .................................................... 7

Practice .............................................................. 8

Word Problem Practice ..................................... 9

Enrichment ...................................................... 10

Lesson 8-2The Pythagorean Theorem and

Its Converse

Study Guide and Intervention .......................... 11

Skills Practice .................................................. 13

Practice .......................................................... 14

Word Problem Practice ................................... 15

Enrichment ...................................................... 16

Spreadsheet Activity ........................................ 17

Lesson 8-3Special Right Triangles


Skills Practice .................................................. 20

Practice .......................................................... 21


Enrichment ...................................................... 23

Lesson 8-4Trigonometry


Skills Practice .................................................. 26

Practice .......................................................... 27


Enrichment ...................................................... 29

Lesson 8-5Angles of Elevation and Depression


Skills Practice .................................................. 32

Practice .......................................................... 33


Enrichment ...................................................... 35

Lesson 8-6The Law of Sines and Law of Cosines


Skills Practice .................................................. 38

Practice .......................................................... 39


Enrichment ...................................................... 41

Graphing Calculator Activity ............................ 42

Lesson 8-7Vectors


Skills Practice .................................................. 45

Practice .......................................................... 46


Enrichment ...................................................... 48

AssessmentStudent Recording Sheet ................................ 49

Rubric for Extended-Response ....................... 50

Chapter 8 Quizzes 1 and 2 ............................. 51

Chapter 8 Quizzes 3 and 4 ............................. 52

Chapter 8 Mid-Chapter Test ............................ 53

Chapter 8 Vocabulary Test ............................. 54

Chapter 8 Test, Form 1 ................................... 55

Chapter 8 Test, Form 2A ................................. 57

Chapter 8 Test, Form 2B ................................. 59

Chapter 8 Test, Form 2C ................................ 61

Chapter 8 Test, Form 2D ................................ 63

Chapter 8 Test, Form 3 ................................... 65

Chapter 8 Extended-Response Test ............... 67

Standardized Test Practice ............................. 68

Answers ........................................... A1–A36

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iv

Teacher’s Guide to Using the

Chapter 8 Resource MastersThe Chapter 8 Resource Masters includes the core materials needed for Chapter 8. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing on the

TeacherWorks PlusTM CD-ROM.

Chapter Resources

Student-Built Glossary (pages 1–2) These

masters are a student study tool that

presents up to twenty of the key vocabulary

terms from the chapter. Students are to

record definitions and/or examples for each

term. You may suggest that students

highlight or star the terms with which they

are not familiar. Give this to students before

beginning Lesson 8-1. Encourage them to

add these pages to their mathematics study

notebooks. Remind them to complete the

appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if

their perceptions have changed.

Lesson Resources

Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator or Spreadsheet

Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

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Assessment OptionsThe assessment masters in the Chapter 8

Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.

Extended–Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level stu-dents. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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NAME DATE PERIOD

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Chapter 8 1 Glencoe Geometry

Student-Built Glossary8

Vocabulary TermFound

on PageDefinition/Description/Example

angle of depression

angle of elevation

component form

cosine

geometric mean

Law of Cosines

Law of Sines

magnitude

(continued on the next page)

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 8.

As you study the chapter, complete each term’s definition or description.

Remember to add the page number where you found the term. Add these pages to

your Geometry Study Notebook to review vocabulary at the end of the chapter.

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Student-Built Glossary (continued)8

Vocabulary TermFound

on PageDefinition/Description/Example

Pythagorean triple

resultant

sine

tangent

trigonometric ratio

trigonometry

vector

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Before you begin Chapter 8

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

Anticipation Guide

Right Triangles and Trigonometry

After you complete Chapter 8

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

8

Step 1

STEP 1A, D, or NS

StatementSTEP 2A or D

1. The geometric mean between two numbers is the positive square root of their product.

2. An altitude drawn from the right angle of a right triangle to its hypotenuse separates the triangle into two congruent triangles.

3. In a right triangle, the length of the hypotenuse is equal to the sum of the lengths of the legs.

4. If any triangle has sides with lengths 3, 4, and 5, then that triangle is a right triangle.

5. If the two acute angles of a right triangle are 45°,

then the length of the hypotenuse is √ " 2 times the

length of either leg.

6. In any triangle whose angle measures are 30°, 60°,

and 90°, the hypotenuse is √ " 3 times as long as the shorter leg.

7. The sine ratio of an angle of a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse.

8. The tangent of an angle of a right triangle whose sides have lengths 3, 4, and 5 will be smaller than the tangent of an angle of a right triangle whose sides have lengths 6, 8, and 10.

9. Trigonometric ratios can be used to solve problems involving angles of elevation and angles of depression.

10. The Law of Sines can only be used in right triangles.

Step 2

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NOMBRE FECHA PERÍODO

Antes de comenzar el Capítulo 8

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a).

PASO 1A, D o NS

EnunciadoPASO 2A o D

1. La media geométrica de dos números es la raíz cuadrada positiva de su producto.

2. Una altitud que se dibuja desde el ángulo recto de un triángulo rectángulo hasta su hipotenusa divide el triángulo en dos triángulos congruentes.

3. En un triángulo rectángulo, la longitud de la hipotenusa es igual a la suma de las longitudes de los catetos.

4. Si un triángulo rectángulo tiene lados con longitudes 3, 4 y 5, entonces cualquier triángulo es un triángulo rectángulo.

5. Si los dos ángulos agudos de un triángulo rectángulo son de 45°, entonces la longitud de la hipotenusa es √ " 2 veces la longitud de cualquiera de los catetos.

6. En cualquier triángulo cuyas medidas de ángulos sean de 30°, 60° y 90°, la hipotenusa es √ " 3 veces tan larga como el cateto más corto.

7. La razón del seno para un ángulo en un triángulo rectángulo es igual a la longitud del lado adyacente dividido entre la longitud de la hipotenusa.

8. La tangente para un ángulo de un triángulo rectángulo cuyos lados tienen longitudes 3, 4 y 5 será menor que la tangente de un ángulo en un triángulo rectángulo cuyos lados tienen longitudes 6, 8 y 10.

9. Las razones trigonométricas se pueden usar para resolver problemas de ángulos de elevación y ángulos de depresión.

10. La ley de los senos sólo se puede usar con triángulos rectángulos.

Después de completar el Capítulo 8

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Ejercicios preparatorios

Triángulos rectángulos y trigonometría

Paso 1

Paso 2

8

Capítulo 8 4 Geometría de Glencoe

Lesso

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Study Guide and Intervention

Geometric Mean

Geometric Mean The geometric mean between two numbers is the positive square root of their product. For two positive numbers a and b, the geometric mean of a and b is

the positive number x in the proportion a −

x =

x −

b . Cross multiplying gives x2 = ab, so x = √ ## ab .

Find the geometric mean between each pair of numbers.

a. 12 and 3

x = √ ## ab Definition of geometric mean

= √ ### 12 . 3 a = 12 and b = 3

= √ ##### (2 . 2 . 3) . 3 Factor.

= 6 Simplify.

The geometric mean between 12 and 3 is 6.

b. 8 and 4

x = √ ## ab Definition of geometric mean

= √ ## 8 . 4 a = 8 and b = 4

= √ #### (2 . 4) . 4 Factor.

= √ ### 16 . 2 Associative Property

= 4 √ # 2 Simplify.

The geometric mean between 8 and 4

is 4 √ # 2 or about 5.7.

Exercises


1. 4 and 4 2. 4 and 6

3. 6 and 9 4. 1 − 2 and 2

5. 12 and 20 6. 4 and 25

7. 16 and 30 8. 10 and 100

9. 1 − 2 and 1 −

4 10. 17 and 3

11. 4 and 16 12. 3 and 24

8-1

Example

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z

y

x

15

R

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S

25

Study Guide and Intervention (continued)

Geometric Mean

Geometric Means in Right Triangles In the diagram, △ ABC ∼ △ ADB ∼ △ BDC. An altitude to the hypotenuse of a right triangle forms two right triangles. The two triangles are similar and each is similar to the original triangle.

Use right △ ABC with

−−

BD ⊥ −−

AC . Describe two geometric

means.

a. △ ADB ∼ △ BDC so AD −

BD = BD

− CD

.

In △ABC, the altitude is the geometric

mean between the two segments of the

hypotenuse.

b. △ ABC ∼ △ ADB and △ ABC ∼ △ BDC,

so AC

− AB

= AB −

AD and

AC −

BC =

BC −

DC .

In △ ABC, each leg is the geometric

mean between the hypotenuse and the

segment of the hypotenuse adjacent to

that leg.

Find x, y, and z.

15 = √ &&&& RP ' SP Geometric Mean (Leg) Theorem

15 = √ && 25x RP = 25 and SP = x

225 = 25x Square each side.

9 = x Divide each side by 25.

Then

y = RP – SP

= 25 – 9

= 16

z = √ &&&& RS ' RP Geometric Mean (Leg) Theorem

= √ &&& 16 ' 25 RS = 16 and RP = 25

= √ && 400 Multiply.

= 20 Simplify.

Exercises

Find x, y, and z to the nearest tenth.

1. x

1 3

2.

z

xy 5

2 3. zx

y

81

4.

xy

1

√3

√"12 5.

x

z y

2

2

6. x z

y

62

8-1

CD

B

A

Example 1 Example 2

Lesso

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Skills Practice

Geometric Mean


1. 2 and 8 2. 9 and 36 3. 4 and 7

4. 5 and 10 5. 28 and 14 6. 7 and 36

Write a similarity statement identifying the three similar triangles in the figure.

7.

C

D

B

A 8.

L

M

N

P

9.

G

E H

F

10.

R T

S

U

Find x, y and z.

11.

3 9

yx

z

12.

10

4

y

x

z

13.

15

4

y

xz

14.

2

5y

x

z

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Practice

Geometric Mean


1. 8 and 12 2. 3 and 15 3. 4 − 5 and 2

Write a similarity statement identifying the three similar triangles in the figure.

4. U

TA

V

5.

KL

J M

Find x, y, and z.

6.

23

z

xy

8 7.

zx

y

625

8.

x

y

2

3

z

9.

x y

10z

20

10. CIVIL An airport, a factory, and a shopping center are at the vertices of a right triangle

formed by three highways. The airport and factory are 6.0 miles apart. Their distances

from the shopping center are 3.6 miles and 4.8 miles, respectively. A service road will be

constructed from the shopping center to the highway that connects the airport and

factory. What is the shortest possible length for the service road? Round to the nearest

hundredth.

8-1

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1. SQUARES Wilma has a rectangle of

dimensions ℓ by w. She would like to

replace it with a square that has the

same area. What is the side length of

the square with the same area as

Wilma’s rectangle?

2. EQUALITY Gretchen computed the

geometric mean of two numbers. One of

the numbers was 7 and the geometric

mean turned out to be 7 as well. What

was the other number?

3. VIEWING ANGLE A photographer

wants to take a picture of a beach front.

His camera has a viewing angle of 90°

and he wants to make sure two palm

trees located at points A and B in the

figure are just inside the edges of the

photograph.

Wal

kway

90 ft 40 ftA B

x

A B

x

He walks out on a walkway that goes

over the ocean to get the shot. If his

camera has a viewing angle of 90°, at

what distance down the walkway should

he stop to take his photograph?

4. EXHIBITIONS A museum has a famous

statue on display. The curator places the

statue in the corner of a rectangular

room and builds a 15-foot-long railing in

front of the statue. Use the information

below to find how close visitors will be

able to get to the statue.

Statue12 ft

15 ft9 ft

x

5. CLIFFS A bridge connects to a tunnel

as shown in the figure. The bridge is

180 feet above the ground. At a distance

of 235 feet along the bridge out of the

tunnel, the angle to the base and

summit of the cliff is a right angle.

Cliff

235 ft

180 f

t

dx

h

a. What is the height of the cliff? Round

to the nearest whole number.

b. How high is the cliff from base to

summit? Round to the nearest whole

number.

c. What is the value of d? Round to the

nearest whole number.

Word Problem Practice

Geometric Mean

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Mathematics and Music

Pythagoras, a Greek philosopher who lived during the sixth century B.C., believed that all nature, beauty, and harmony could be expressed by whole-number relationships. Most people remember Pythagoras for his teachings about right triangles. (The sum of the squares of the legs equals the square of the hypotenuse.) But Pythagoras also discovered relationships between the musical notes of a scale. These relationships can be expressed as ratios.

C D E F G A B C′1 −

1

8 −

9 4

− 5

3 −

4 2

− 3

3 −

5

8 −

15 1

− 2

When you play a stringed instrument, The C string can be usedyou produce different notes by placing to produce F by placing

your finger on different places on a string. a finger 3 −

4 of the way

This is the result of changing the length along the string.of the vibrating part of the string.

Suppose a C string has a length of 16 inches. Write and solve

proportions to determine what length of string would have to

vibrate to produce the remaining notes of the scale.

1. D 2. E 3. F

4. G 5. A 6. B

7. C′

8. Complete to show the distance between finger positions on the 16-inch

C string for each note. For example, C(16) - D (14 2 − 9 ) = 1

7 −

9 .

C D E F G A B C′

34

of C string

Enrichment

1 7−9

in.

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Less

on

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The Pythagorean Theorem and Its Converse

The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of

the hypotenuse. If the three whole numbers a, b, and c satisfy the equation

a2 + b2 = c2, then the numbers a, b, and c form a

Pythagorean triple.

ca

bA C

B

a. Find a.

a

12

13

AC

B

a2 + b2 = c2 Pythagorean Theorem

a2 + 122 = 132 b = 12, c = 13

a2 + 144 = 169 Simplify.

a2 = 25 Subtract.

a = 5 Take the positive square root

of each side.

b. Find c.

c

30

20

AC

B

a2 + b2 = c2 Pythagorean Theorem

202 + 302 = c2 a = 20, b = 30

400 + 900 = c2 Simplify.

1300 = c2 Add.

√ "" 1300 = c Take the positive square root

of each side.

36.1 ≈ c Use a calculator.

Find x.

1.

x

3 3 2.

x

159

3.

x

6525

4. x

5

9

4

9 5.

x

33

16

6. x

1128

Use a Pythagorean Triple to find x.

7. 8

17

x

8. 45

24x

9. 28

96

x

8-2

Example

Exercises

△ ABC is a right triangle.

so a2 + b2 = c2.

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c

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B



Converse of the Pythagorean Theorem If the sum of the squares of the lengths of the two shorter sides of a triangle equals the square of the lengths of the longest side, then the triangle is a right triangle.

You can also use the lengths of sides to classify a triangle. If a2 + b2 = c2, then

if a2 + b2 = c2 then △ABC is a right triangle. △ABC is a right triangle.

if a2 + b2 > c2 then △ABC is acute. if a2 + b2 < c2 then △ABC is obtuse.

Determine whether △PQR is a right triangle.

a2 + b2 " c2 Compare c2 and a2 + b2

102 + (10 √ $ 3 )2 " 202 a = 10, b = 10 √ $ 3 , c = 20

100 + 300 " 400 Simplify.

400 = 400% Add.

Since c2 = and a2 + b2, the triangle is a right triangle.

Exercises

Determine whether each set of measures can be the measures of the sides of a

triangle. If so, classify the triangle as acute, obtuse, or right. Justify your answer.

1. 30, 40, 50 2. 20, 30, 40 3. 18, 24, 30

4. 6, 8, 9 5. 6, 12, 18 6. 10, 15, 20

7. √ $ 5 , √ $$ 12 , √ $$ 13 8. 2, √ $ 8 , √ $$ 12 9. 9, 40, 41

2010

10√3

8-2

Example

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Skills Practice


Find x.

1. x

12

9

2. x

12

13 3.

x

12

32

4. x12.5

25

5.

x

9 9

8

6.

x

31

14


7.

5

12

x

8.

8

10

x 9.

20

12

x

10.

65

25

x

11. 12.

50

48

x

Determine whether each set of numbers can be measure of the sides of a triangle.

If so, classify the triangle as acute, obtuse, or right. Justify your answer.

13. 7, 24, 25 14. 8, 14, 20 15. 12.5, 13, 26

16. 3 √ " 2 , √ " 7 , 4 17. 20, 21, 29 18. 32, 35, 70

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Practice


Find x.

1. x

13

23

2.

x

3421

3. x

26

2618

4.

x

34

22

5. x

16

14

6. x

2424

42


7.

36

27

x

8.

120

136

x

9.

65

39

x

10. 42

150

x

Determine whether each set of numbers can be measure of the sides of a triangle.

If so, classify the triangle as acute, obtuse, or right. Justify your answer.

11. 10, 11, 20 12. 12, 14, 49 13. 5 √ " 2 , 10, 11

14. 21.5, 24, 55.5 15. 30, 40, 50 16. 65, 72, 97

17. CONSTRUCTION The bottom end of a ramp at a warehouse is

10 feet from the base of the main dock and is 11 feet long. How

high is the dock?

11 ft?

dock

ramp

10 ft

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1. SIDEWALKS Construction workers are building a marble sidewalk around a park that is shaped like a right triangle. Each marble slab adds 2 feet to the length of the sidewalk. The workers find that exactly 1071 and 1840 slabs are required to make the sidewalks along the short sides of the park. How many slabs are required to make the sidewalk that runs along the long side of the park?

2. RIGHT ANGLES Clyde makes a triangle using three sticks of lengths 20 inches, 21 inches, and 28 inches. Is the triangle a right triangle? Explain.

3. TETHERS To help support a flag pole, a 50-foot-long tether is tied to the pole at a point 40 feet above the ground. The tether is pulled taut and tied to an anchor in the ground. How far away from the base of the pole is the anchor?

4. FLIGHT An airplane lands at an airport 60 miles east and 25 miles north of where it took off.

60 mi

25 mi

How far apart are the two airports?

5. PYTHAGOREAN TRIPLES Ms. Jones assigned her fifth-period geometry class the following problem.

Let m and n be two positive integers with m > n. Let a = m2 – n2, b = 2mn, and c = m2 + n2.

a. Show that there is a right triangle with side lengths a, b, and c.

b. Complete the following table.

m n a b c

2 1 3 4 5

3 1

3 2

4 1

4 2

4 3

5 1

c. Find a Pythagorean triple that corresponds to a right triangle with a hypotenuse 252 = 625 units long. (Hint: Use the table you completed for Exercise b to find two positive integers m and n with m > n and m2 + n2 = 625.)

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Enrichment

Converse of a Right Triangle TheoremYou have learned that the measure of the altitude from the vertex of

the right angle of a right triangle to its hypotenuse is the geometric

mean between the measures of the two segments of the hypotenuse.

Is the converse of this theorem true? In order to find out, it will help

to rewrite the original theorem in if-then form as follows.

If △ABQ is a right triangle with right angle at Q, then

QP is the geometric mean between AP and PB, where P

is between A and B and −−−

QP is perpendicular to −−

AB .

1. Write the converse of the if-then form of the theorem.

2. Is the converse of the original theorem true? Refer

to the figure at the right to explain your answer.

You may find it interesting to examine the other theorems in

Chapter 8 to see whether their converses are true or false. You will

need to restate the theorems carefully in order to write their

converses.

Q

BP

A

Q

BP

A

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Spreadsheet Activity

Pythagorean Triples

You can use a spreadsheet to determine whether three whole numbers form a Pythagorean triple.

Use a spreadsheet to determine whether the numbers 12, 16, and 20

form a Pythagorean triple.

Step 1 In cell A1, enter 12. In cell B1, enter 16 and in cell C1, enter 20. The longest side should be entered in column C.

Step 2 In cell D1, enter an equals sign followed by IF(A1^2+B1^2=C1^2,“YES”,“NO”). This will return “YES” if the set of numbers is a Pythagorean triple and will return “NO” if itis not.

The numbers 12, 16, and 20 form a Pythagorean triple.

Use a spreadsheet to determine whether the numbers 3, 6, and 12

form a Pythagorean triple.

Step 1 In cell A2, enter 3, in cell B2, enter 6, and in cell C2, enter 12.

Step 2 Click on the bottom right corner of cell D1 and drag it to D2. This will determine whether or not the set of numbers is a Pythagorean triple.

The numbers 3, 6, and 12 do not form a Pythagorean triple.

Exercises

Use a spreadsheet to determine whether each set of numbers

forms a Pythagorean triple.

1. 14, 48, 50 2. 16, 30, 34 3. 5, 5, 9

4. 4, 5, 7 5. 18, 24, 30 6. 10, 24, 26

7. 25, 60, 65 8. 2, 4, 5 9. 19, 21, 22

10. 18, 80, 82 11. 5, 12, 13 12. 20, 48, 52

8-2

Example 1

Example 2

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Special Right Triangles

Properties of 45°-45°-90° Triangles The sides of a 45°-45°-90° right triangle have a special relationship.

If the leg of a 45°-45°-90°

right triangle is x units, show

that the hypotenuse is x √ " 2 units.

x √

x

x245°

45°

Using the Pythagorean Theorem with

a = b = x, then

c2 = a2 + b2

c2 = x2 + x2

c2 = 2x2

c = √ ## 2x2

c = x √ # 2

In a 45°-45°-90° right

triangle the hypotenuse is √ " 2 times

the leg. If the hypotenuse is 6 units,

find the length of each leg.

The hypotenuse is √ # 2 times the leg, so

divide the length of the hypotenuse by √ # 2 .

a = 6 −

√ # 2

= 6 −

√ # 2 .

√ # 2 −

√ # 2

= 6 √ # 2

− 2

= 3 √ # 2 units

Exercises

Find x.

1. x

8

45°

45°

2.

x

45°3√2 3.

45°

4

x

4.

x x

18 5.

45°

16

x

x

6.

45°

24

x

√2

7. If a 45°-45°-90° triangle has a hypotenuse length of 12, find the leg length.

8. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of

25 inches.

9. Find the length of the hypotenuse of a 45°-45°-90° triangle with a leg length of

14 centimeters.

8-3

Example 1 Example 2

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Properties of 30°-60°-90° Triangles The sides of a 30°-60°-90° right triangle also have a special relationship.

In a 30°-60°-90° right triangle the hypotenuse

is twice the shorter leg. Show that the longer leg is √ " 3 times

the shorter leg.

△ MNQ is a 30°-60°-90° right triangle, and the length of the hypotenuse

−−− MN is two times the length of the shorter side

−−− NQ .

Use the Pythagorean Theorem.

a2 = (2x)2 - x2 a2 = c2 - b2

a2 = 4x2 - x2 Multiply.

a2 = 3x2 Subtract.

a = √ $$ 3x2 Take the positive square root of each side.

a = x √ $ 3 Simplify.

In a 30°-60°-90° right triangle, the hypotenuse is 5 centimeters.

Find the lengths of the other two sides of the triangle.

If the hypotenuse of a 30°-60°-90° right triangle is 5 centimeters, then the length of the

shorter leg is one-half of 5, or 2.5 centimeters. The length of the longer leg is √ $ 3 times the

length of the shorter leg, or (2.5)( √ $ 3 ) centimeters.

Exercises

Find x and y.

1. x

y30°

60°1

2

2.

x

y

60°

8

3.

x

y

11

30°

4.

xy

30°

9√3

5.

xy60°

12

6. 60°

x

20y

7. An equilateral triangle has an altitude length of 36 feet. Determine the length of a side of the triangle.

8. Find the length of the side of an equilateral triangle that has an altitude length of 45 centimeters.

8-3

30°

60°

x2x

aExample 1

Example 2

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8-3 Skills Practice


Find x.

1.

45°

25

x

2.

45°

17x

3. 45°

48

x

4.

45°

100

x

5.

45°

100

x

6.

45°

88 x

7. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of 26.


50 centimeters.

Find x and y.

9.

30°

x11

y

10.

60°x

8

y

√3 11. 30°

x

5

y

√3

12.

60°

x

30

y

13.

30°

x

21y

√3

14. 60°

x

52y

√3

15. An equilateral triangle has an altitude length of 27 feet. Determine the length of a side

of the triangle.

16. Find the length of the side of an equilateral triangle that has an altitude length of

11 √ " 3 feet.

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Find x.

1.

45°

14

x

2.

45°

45x

3. 45°

22

x

4.

45°

210

x

5.

45°

88

x

6.

45°

x5√2

Find x and y.

7.

30°

x9

y

8.

60°x

4

y

√3

9.

30°

x

20

y

10.

60°

x

98

y

11. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of 38.


77 centimeters.

13. An equilateral triangle has an altitude length of 33 feet. Determine the length of a side

of the triangle.

14. BOTANICAL GARDENS One of the displays at a botanical garden

is an herb garden planted in the shape of a square. The square

measures 6 yards on each side. Visitors can view the herbs from a

diagonal pathway through the garden. How long is the pathway?6 yd 6 yd

6 yd

6 yd

Practice


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1. ORIGAMI A square piece of paper

150 millimeters on a side is folded in

half along a diagonal. The result is a

45°-45°-90° triangle. What is the length

of the hypotenuse of this triangle?

2. ESCALATORS A 40-foot-long escalator

rises from the first floor to the second

floor of a shopping mall. The escalator

makes a 30° angle with the horizontal.

40 ft

30°

? ft

How high above the first floor is the

second floor?

3. HEXAGONS A box of chocolates shaped

like a regular hexagon is placed snugly

inside of a rectangular box as shown in

the figure.

If the side length of the hexagon is

3 inches, what are the dimensions of

the rectangular box?

4. WINDOWS A large stained glass

window is constructed from six 30°-60°-

90° triangles as shown in the figure.

3 m

What is the height of the window?

5. MOVIES Kim and Yolanda are

watching a movie in a movie theater.

Yolanda is sitting x feet from the screen

and Kim is 15 feet behind Yolanda.

30° 45°

15 ft x ft

The angle that Kim’s line of sight

to the top of the screen makes with

the horizontal is 30°. The angle that

Yolanda’s line of sight to the top of the

screen makes with the horizontal is 45°.

a. How high is the top of the screen in

terms of x?

b. What is x + 15

− x ?

c. How far is Yolanda from the screen?

Round your answer to the nearest

tenth.

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Enrichment

√

1

1

1

3√

2

Constructing Values of Square RootsThe diagram at the right shows a right isosceles triangle with

two legs of length 1 inch. By the Pythagorean Theorem, the length

of the hypotenuse is √ # 2 inches. By constructing an adjacent right

triangle with legs of √ # 2 inches and 1 inch, you can create a segment

of length √ # 3 .

By continuing this process as shown below, you can construct a

“wheel” of square roots. This wheel is called the “Wheel of Theodorus”

after a Greek philosopher who lived about 400 B.C.

Continue constructing the wheel until you make a segment of

length √ ## 18 .

1

1

11

1

1

√2

√3

√5

√6

√4 = 2

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Trigonometric Ratios The ratio of the lengths of two sides of a right triangle is called a trigonometric ratio. The three most common ratios are sine, cosine, and tangent, which are abbreviated sin, cos, and tan, respectively.

sin R = leg opposite ∠R

− hypotenuse

cos R = leg adjacent to ∠R

−− hypotenuse

tan R = leg opposite ∠R

−− leg adjacent to ∠R

= r − t =

s −

t =

r −

s

Find sin A, cos A, and tan A. Express each ratio as

a fraction and a decimal to the nearest hundredth.

sin A = opposite leg

− hypotenuse

cos A = adjacent leg

− hypotenuse

tan A = opposite leg

− adjacent leg

= BC

− BA

= AC

− AB

= BC

− AC

= 5 −

13 = 12

− 13

= 5 −

12

≈ 0.38 ≈ 0.92 ≈ 0.42

Exercises

Find sin J, cos J, tan J, sin L, cos L, and tan L. Express each ratio as a fraction

and as a decimal to the nearest hundredth if necessary.

1. 2. 3.

12

135

C

B

A


Trigonometry

8-4

s

tr

T

S

R

Example

20

12

16

40

24

32

36

12√324√3

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8-4



Trigonometry

Use Inverse Trigonometric Ratios You can use a calculator and the sine, cosine, or tangent to find the measure of the angle, called the inverse of the trigonometric ratio.

Use a calculator to find the measure of ∠T to the nearest tenth.

The measures given are those of the leg opposite ∠T and the

hypotenuse, so write an equation using the sine ratio.

sin T = opp

− hyp

sin T = 29 −

34

If sin T = 29

− 34

, then sin-1 29

− 34

= m∠T.

Use a calculator. So, m∠T ≈ 58.5.

Exercises

Use a calculator to find the measure of ∠T to the nearest tenth.

1.

34

14√3

2. 7

18

3.

87

34

4.

101

32

5.

67

10√3

6.

39

14√2

8-4

Example

34

29

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Find sin R, cos R, tan R, sin S, cos S, and tan S.

Express each ratio as a fraction and as a decimal to the

nearest hundredth.

1. r = 16, s = 30, t = 34 2. r = 10, s = 24, t = 26

Use a special right triangle to express each trigonometric ratio as a fraction and

as a decimal to the nearest hundredth if necessary.

3. sin 30° 4. tan 45° 5. cos 60°

6. sin 60° 7. tan 30° 8. cos 45°

Find x. Round to the nearest hundredth if necessary.

9.

7 x

36°

10. 11.

Use a calculator to find the measure of ∠B to the nearest tenth.

12. 13. 14.

sR

S

T

rt

Skills Practice

Trigonometry

8-4

12x

15°

6

x°

18

15

x°

12

22

x°

19

12 x

63°

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Find sin L, cos L, tan L, sin M, cos M, and tan M.

Express each ratio as a fraction and as a decimal to the

nearest hundredth.

1. ℓ = 15, m = 36, n= 39 2. ℓ = 12, m = 12 √ # 3 , n = 24

Find x. Round to the nearest hundredth.

3.

11

64°

x

4. 5.

41°

x

32

Use a calculator to find the measure of ∠B to the nearest tenth.

6.

8

14

7.

30

25

8.

9. GEOGRAPHY Diego used a theodolite to map a region of land for his

class in geomorphology. To determine the elevation of a vertical rock

formation, he measured the distance from the base of the formation to

his position and the angle between the ground and the line of sight to the

top of the formation. The distance was 43 meters and the angle was

36°. What is the height of the formation to the nearest meter?

29

29°x

36°

43m

ML

N

m

n

Practice

Trigonometry

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Trigonometry

1. RADIO TOWERS Kay is standing near

a 200-foot-high radio tower.

200 ft ? ft

49°

Use the information in the figure to

determine how far Kay is from the top of

the tower. Express your answer as a

trigonometric function.

2. RAMPS A 60-foot ramp rises from the

first floor to the second floor of a parking

garage. The ramp makes a 15° angle

with the ground.

? ft

60 ft

15°

How high above the first floor is the

second floor? Express your answer as a

trigonometric function.

3. TRIGONOMETRY Melinda and Walter

were both solving the same trigonometry

problem. However, after they finished

their computations, Melinda said the

answer was 52 sin 27° and Walter said

the answer was 52 cos 63°. Could they

both be correct? Explain.

4. LINES Jasmine draws line m on a

coordinate plane.

m

x

y

O-5

5

5

What angle does m make with the

x-axis? Round your answer to the

nearest degree.

5. NEIGHBORS Amy, Barry, and Chris live

on the same block. Chris lives up the

street and around the corner from Amy,

and Barry lives at the corner between

Amy and Chris. The three homes are the

vertices of a right triangle.

64°

Barry

Chris

Amy

26°

a. Give two trigonometric expressions

for the ratio of Barry’s distance from

Amy to Chris’ distance from Amy.

b. Give two trigonometric expressions

for the ratio of Barry’s distance from

Chris to Amy’s distance from Chris.

c. Give a trigonometric expression for

the ratio of Amy’s distance from

Barry to Chris’ distance from Barry.

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Enrichment

Sine and Cosine of AnglesThe following diagram can be used to obtain approximate values for the sine and cosine of angles from 0° to 90°. The radius of the circle is 1. So, the sine and cosine values can be read directly from the vertical and horizontal axes.

Find approximate values for sin 40° and

cos 40°. Consider the triangle formed by the segment

marked 40°, as illustrated by the shaded triangle at right.

sin 40° = a −

c ≈

0.64 −

1 or 0.64 cos 40° =

b −

c ≈

0.77 −

1 or 0.77

1. Use the diagram above to complete the chart of values.

x° 0° 10° 20° 30° 40° 50° 60° 70° 80° 90°

sin x° 0.64

cos x° 0.77

2. Compare the sine and cosine of two complementary angles (angles with a sum of 90°). What do you notice?

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

90°

0°

10°

20°

30°

40°

50°

60°

70°

80°

1

0

40°0.64

c = 1 unit

x°

b = cos x° 0.77 1

a = sin x°

Example

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Angles of Elevation and Depression

Angles of Elevation and Depression Many real-world problems that involve looking up to an object can be described in terms of an angle of elevation, which is the angle between an observer’s line of sight and a horizontal line.

When an observer is looking down, the angle of depression is the angle between the observer’s line of sight and a horizontal line.

The angle of elevation from point A to the top

of a cliff is 34°. If point A is 1000 feet from the base of the cliff,

how high is the cliff?

Let x = the height of the cliff.

tan 34° = x −

1000 tan =

opposite −

adjacent

1000(tan 34°) = x Multiply each side by 1000.

674.5 ≈ x Use a calculator.

The height of the cliff is about 674.5 feet.

Exercises

1. HILL TOP The angle of elevation from point A to the top of a hill is 49°. If point A is 400 feet from the base of the hill, how high is the hill?

2. SUN Find the angle of elevation of the Sun when a 12.5-meter-tall telephone pole casts an 18-meter-long shadow.

3. SKIING A ski run is 1000 yards long with a vertical drop of 208 yards. Find the angle of depression from the top of the ski run to the bottom.

4. AIR TRAFFIC From the top of a 120-foot-high tower, an air traffic controller observes an airplane on the runway at an angle of depression of 19°. How far from the base of the tower is the airplane?

18 m

12.5 m

Sun

?

400 ft

?

49°A

?

1000 ft

34°

x

angle ofelevation

line o

f sight

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Example

Yline of s

ight

horizontal

angle ofdepression

120 ft

?

19°

208 yd

?

1000 yd

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Two Angles of Elevation or Depression Angles of elevation or depression to two different objects can be used to estimate distance between those objects. The angles from two different positions of observation to the same object can be used to estimate the height of the object.

To estimate the height of a garage,

Jason sights the top of the garage at a 42° angle

of elevation. He then steps back 20 feet and sites

the top at a 10° angle. If Jason is 6 feet tall,

how tall is the garage to the nearest foot?

△ ABC and △ ABD are right triangles. We can determine AB = x and CB = y, and DB = y + 20.

Use △ ABC. Use △ ABD.

tan 42° = x − y or y tan 42° = x tan 10° =

x −

y + 20 or (y + 20) tan 10° = x

Substitute the value for x from △ ABD in the equation for △ ABC and solve for y.

y tan 42° = (y + 20) tan 10°

y tan 42° = y tan 10° + 20 tan 10°

y tan 42° − y tan 10° = 20 tan 10°

y (tan 42° − tan 10°) = 20 tan 10°

y = 20 tan 10°

−− tan 42° - tan 10°

≈ 4.87

If y ≈ 4.87, then x = 4.87 tan 42° or about 4.4 feet. Add Jason’s height, so the garage is about 4.4 + 6 or 10.4 feet tall.

Exercises

1. CLIFF Sarah stands on the ground and sights the top of a steep cliff at a 60° angle of elevation. She then steps back 50 meters and sights the top of the steep cliff at a 30° angle. If Sarah is 1.8 meters tall, how tall is the steep cliff to the nearest meter?

2. BALLOON The angle of depression from a hot air balloon in the air to a person on the ground is 36°. If the person steps back 10 feet, the new angle of depression is 25°. If the person is 6 feet tall, how far off the ground is the hot air balloon?



8-5

Example

42°10°

x

y6 ft 20 ft

Building

60°30°

x

y1.8m 50m

Steepcliff

36°25°

x

y6 ft 10 ft

Balloon

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Name the angle of depression or angle of elevation in each figure.

1. 2.

3. 4.

5. MOUNTAIN BIKING On a mountain bike trip along the Gemini Bridges Trail in Moab,

Utah, Nabuko stopped on the canyon floor to get a good view of the twin sandstone

bridges. Nabuko is standing about 60 meters from the base of the canyon cliff, and the

natural arch bridges are about 100 meters up the canyon wall. If her line of sight is

5 metres above the ground, what is the angle of elevation to the top of the bridges?

Round to the nearest tenth degree.

6. SHADOWS Suppose the sun casts a shadow off a 35-foot building.

If the angle of elevation to the sun is 60°, how long is the shadow

to the nearest tenth of a foot?

7. BALLOONING Angie sees a hot air balloon in the

sky from her spot on the ground. The angle of

elevation from Angie to the balloon is 40°. If she steps

back 200 feet, the new angle of elevation is 10°.

If Angie is 5.5 feet tall, how far off the ground

is the hot air balloon?

8. INDIRECT MEASUREMENT Kyle is at the end

of a pier 30 feet above the ocean. His eye level is

3 feet above the pier. He is using binoculars to

watch a whale surface. If the angle of depression

of the whale is 20°, how far is the whale from

Kyle’s binoculars? Round to the nearest tenth foot.

whale water level

20°

Kyle’s eyes

pier3 ft

30 ft

60°?

35 ft

Z

P

W

R

D

A

C

B

T

W

R

S

F

T

L

S

Skills Practice


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40°

10°x

y5.5 ft 200 ft

Balloon

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Name the angle of depression or angle of elevation in each figure.

1. 2.

3. WATER TOWERS A student can see a water tower from the closest point of the soccer

field at San Lobos High School. The edge of the soccer field is about 110 feet from the

water tower and the water tower stands at a height of 32.5 feet. What is the angle of

elevation if the eye level of the student viewing the tower from the edge of the soccer

field is 6 feet above the ground? Round to the nearest tenth.

4. CONSTRUCTION A roofer props a ladder against a wall so that the top of the ladder

reaches a 30-foot roof that needs repair. If the angle of elevation from the bottom of the

ladder to the roof is 55°, how far is the ladder from the base of the wall? Round your

answer to the nearest foot.

5. TOWN ORDINANCES The town of Belmont restricts the height

of flagpoles to 25 feet on any property. Lindsay wants to

determine whether her school is in compliance with the regulation.

Her eye level is 5.5 feet from the ground and she stands 36 feet

from the flagpole. If the angle of elevation is about 25°,

what is the height of the flagpole to the nearest tenth?

6. GEOGRAPHY Stephan is standing on the ground by

a mesa in the Painted Desert. Stephan is 1.8 meters

tall and sights the top of the mesa at 29°. Stephan steps

back 100 meters and sights the top at 25°. How tall is

the mesa?

7. INDIRECT MEASUREMENT Mr. Dominguez is standing

on a 40-foot ocean bluff near his home. He can see his two

dogs on the beach below. If his line of sight is 6 feet above

the ground and the angles of depression to his dogs are

34° and 48°, how far apart are the dogs to the nearest foot?

48°34°

40 ft

6 ft

Mr. Dominguez

bluff

25°

5.5 ft

36 ft

x

R

M

P

L

T

Y

R

Z Y

Practice


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29°25°

x

y1.8 m

100 m

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1. LIGHTHOUSES Sailors on a ship at sea

spot the light from a lighthouse. The

angle of elevation to the light is 25°.

25°

30m

?

The light of the lighthouse is 30 meters

above sea level. How far from the shore

is the ship? Round your answer to the

nearest meter.

2. RESCUE A hiker dropped his backpack

over one side of a canyon onto a ledge

below. Because of the shape of the cliff,

he could not see exactly where it landed.

32°

115 ft

? ft

Backpack

From the other side, the park ranger

reports that the angle of depression to

the backpack is 32°. If the width of the

canyon is 115 feet, how far down did

the backpack fall? Round your answer

to the nearest foot.

3. AIRPLANES The angle of elevation to

an airplane viewed from the control

tower at an airport is 7°. The tower is

200 feet high and the pilot reports that

the altitude is 5200 feet. How far away

from the control tower is the airplane?

Round your answer to the nearest foot.

4. PEAK TRAM The Peak Tram in Hong

Kong connects two terminals, one at the

base of a mountain, and the other at the

summit. The angle of elevation of the

upper terminal from the lower terminal

is about 15.5°. The distance between the

two terminals is about 1365 meters.

About how much higher above sea level

is the upper terminal compared to the

lower terminal? Round your answer to

the nearest meter.

5. HELICOPTERS Jermaine and John are

watching a helicopter hover above the

ground.

48° 55°

(Not drawn to scale)

Helicopter

Jermaine John10 m

h

x

Jermaine and John are standing

10 meters apart.

a. Find two different expressions that

can be used to find the h, height of

the helicopter.

b. Equate the two expressions you found

for Exercise a to solve for x. Round

your answer to the nearest

hundredth.

c. How high above the ground is the

helicopter? Round your answer to the

nearest hundredth.

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Enrichment

Best Seat in the House

Most people want to sit in the best seat in the movie theater. The best seat could be

defined as the seat that allows you to see the maximum amount of screen. The picture

below represents this situation.

To determine the best seat in the house, you want to find what value of x allows you to see

the maximum amount of screen. The value of x is how far from the screen you should sit.

1. To maximize the amount of screen viewed, which angle value needs to be maximized?

Why?

2. What is the value of a if x = 10 feet?





7. Which value of x gives the greatest value of a? So, where is the best seat in the movie

theater?

a˚

b˚

c˚

x ft

screen

40 ft

10 ft

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The Law of Sines and Law of Cosines

The Law of Sines In any triangle, there is a special relationship between the angles of the triangle and the lengths of the sides opposite the angles.

Find b. Round to the

nearest tenth.

45°

3074°

b

B

AC

sin C

− c =

sin B −

b Law of Sines

sin 45°

− 30

= sin 74°

− b m∠C = 45, c = 30, m∠B = 74

b sin 45° = 30 sin 74° Cross Products Property

b = 30 sin 74°

− sin 45°

Divide each side by sin 45°.

b ≈ 40.8 Use a calculator.

Find d. Round to the

nearest tenth.

By the Triangle Angle-Sum Theorem, m∠E = 180 - (82 + 40) or 58.

82° 40°

24

d

sin D

− d =

sin E −

e Law of Sines

sin 82°

− d =

sin 58° −

24 m∠D = 82, m∠E = 58,

e = 24

24 sin 82° = d sin 58° Cross Products Property

24 sin 82° −

sin 58° = d Divide each side by sin 58°.

d ≈ 28.0 Use a calculator.

Exercises

Find x. Round to the nearest tenth.

1.

80°

40°

x

12

2.

52°

91°

x

20 3.

16x

84°

34°

4.

17° 72°

x25

5.

89°

80°x

12 6.

60°

35°

35 x

8-6

Example 1 Example 2

Law of Sines sin A

− a =

sin B −

b =

sin C −

c

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The Law of Cosines Another relationship between the sides and angles of any triangle is called the Law of Cosines. You can use the Law of Cosines if you know three sides of a triangle or if you know two sides and the included angle of a triangle.

Law of Cosines

Let △ABC be any triangle with a, b, and c representing the measures of the sides opposite

the angles with measures A, B, and C, respectively. Then the following equations are true.

a2 = b2 + c2 - 2bc cos A b2 = a2 - c2 - 2ac cos B c2 = a2 + b2 - 2ab cos C

Find c. Round to the nearest tenth.

c2 = a2 + b2 - 2ab cos C Law of Cosines

c2 = 122 + 102 - 2(12)(10)cos 48° a = 12, b = 10, m∠C = 48

c = √ %%%%%%%%%%%% 122 + 102 - 2(12)(10)cos 48° Take the square root of each side.

c ≈ 9.1 Use a calculator.

Find m∠A. Round to the nearest degree.

a2 = b2 + c2 - 2bc cos A Law of Cosines

72 = 52 + 82 - 2(5)(8) cos A a = 7, b = 5, c = 8

49 = 25 + 64 - 80 cos A Multiply.

-40 = -80 cos A Subtract 89 from each side.

1 − 2 = cos A Divide each side by -80.

cos-1 1 − 2 = A Use the inverse cosine.

60° = A Use a calculator.

Exercises

Find x. Round angle measures to the nearest degree and side measures to the

nearest tenth.

1. 2. 3.

4. 5. 6.

8-6

Example 1

Example 2

48°10 12

c

C

BA

12

14 x

62°

12

11

10

x°

16

24

18

x°

25

20x

82°

18

28

x

59°

18

15

15

x°

8

5

7

C

B

A

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Find x. Round angle measures to the nearest degree and side lengths to the

nearest tenth.

1.

28

35° 48°

x

2.

18

46°

17°

x

3.

38

x

86°

51°

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. Solve the triangle. Round angle measures to the nearest degree.

Skills Practice


8-6

x8

73°

80°

x

34

41°

88°

x

1283°

60°

18

x

104°

22°

x 27

54°

93°

12 16

11

x°

17

18

23

29

111°

34

x

16

89°

12

x 13

103°

20

x

13

35

x

96° 28

x°26

30

25

28°

x20

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Practice


13. INDIRECT MEASUREMENT To find the distance from the edge

of the lake to the tree on the island in the lake, Hannah set up

a triangular configuration as shown in the diagram. The distance

from location A to location B is 85 meters. The measures of the

angles at A and B are 51° and 83°, respectively. What is the

distance from the edge of the lake at B to the tree on the island at C?

A

C

B

Find x. Round angle measures to the nearest degree and side lengths to the

nearest tenth.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

14

x

67°

14°

127

x

42°

61°

x

5.8

83°

73°

19.1

x

56°

34°

9.6

x

43°

123° 4.3

x

85°

64°

40

12

37°

x 23

28

37°

x

15

17

62°

x

28.414.5

21.7

x°

89°

14

15

x

1410

9.6

x°

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1. ALTITUDES In triangle ABC, the

altitude to side −−

AB is drawn.

A

ab

B

C

Give two expressions for the length of

the altitude in terms of a, b, and the

sine of the angles A and B.

2. MAPS Three cities form the vertices of

a triangle. The angles of the triangle are

40°, 60°, and 80°. The two most distant

cities are 40 miles apart. How close are

the two closest cities? Round your

answer to the nearest tenth of a mile.

3. STATUES Gail was visiting an art

gallery. In one room, she stood so that

she had a view of two statues, one of a

man, and the other of a woman. She was

40 feet from the statue of the woman,

and 35 feet from the statue of the man.

The angle created by the lines of sight to

the two statues was 21°. What is the

distance between the two statues?

Round your answer to the nearest tenth.

4. CARS Two cars start moving from the

same location. They head straight, but

in different directions. The angle

between where they are heading is 43°.

The first car travels 20 miles and the

second car travels 37 miles. How far

apart are the two cars? Round your

answer to the nearest tenth.

5. ISLANDS Oahu

is a Hawaiian

Island. Off of the

coast of Oahu,

there is a very

tiny island

known as

Chinaman’s Hat.

Keoki and Malia

are observing

Chinaman’s Hat from locations

5 kilometers apart. Use the information

in the figure to answer the following

questions.

a. How far is Keoki from Chinaman’s

Hat? Round your answer to the

nearest tenth of a kilometer.

b. How far is Malia from Chinaman’s

Hat? Round your answer to the

nearest tenth of a kilometer.

Ka’a’awa

Keoki Chinaman’s

Hat

5 km

Malia

Kahalu’u

22.3°

49.5°

8-6

Statue of

a man

Statue of a woman

35 ft

40 ft

21˚

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Enrichment

Identities

An identity is an equation that is true for all values of the variable for which both sides are defined. One way to verify an identity is to use a right triangle and the definitions for trigonometric functions.

Verify that (sin A)2 + (cos A)2 = 1 is an identity.

(sin A)2 + (cos A)2 = ( a − c ) 2 + (

b − c ) 2

= a2+ b2

− c2

= c2

− c2

= 1

To check whether an equation may be an identity, you can test several values. However, since you cannot test all values, you cannot be certain that the equation is an identity.

Test sin 2x = 2 sin x cos x to see if it could be an identity.

Try x = 20. Use a calculator to evaluate each expression.

sin 2x = sin 40 2 sin x cos x = 2 (sin 20)(cos 20) ≈ 0.643 ≈ 2(0.342)(0.940) ≈ 0.643

Since the left and right sides seem equal, the equation may be an identity.

Exercises

Use triangle ABC shown above. Verify that each equation is an identity.

1. cos A

− sin A

= 1 −

tan A 2.

tan B −

sin B = 1

− cos B

3. tan B cos B = sin B 4. 1 - (cos B)2 = (sin B)2

Try several values for x to test whether each equation could be an identity.

5. cos 2x = (cos x)2 - (sin x)2 6. cos (90 - x) = sin x

B

A C

ca

b

8-6

Example 1

Example 2

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Graphing Calculator Activity

Solving Triangles Using the Law of Sines or Cosines

You can use a calculator to solve triangles using the Law of Sines or Cosines.

Solve △ABC if a = 6, b = 2, and c = 7.5.

Use the Law of Cosines.

a2 = b2 + c2 - 2bc cos A

62 = 22 + 7.52 - 2(2)(7.5) cos A

m∠A = cos-1 62 - 22 - 7.52

− -2(2)(7.5)

Use your calculator to find the measure of ∠A.

Keystrokes: 2nd [COS-1] ( 6 x2 — 2 x2 — 7.5 x2 ) ÷

Enter: ( (–) 2 3 2 3 7.5 ) ) ENTER 36.06658826

So m∠A ≈ 36. Use the Law of Sines and your calculator to find m∠B.

sin A

− a =

sin B −

b

sin 36

− 6 ≈

sin B −

2

m∠B ≈ sin-1 2 sin 36°

− 6

Keystrokes: 2nd [SIN-1] ( 2 SIN 36 ) ÷ 6 ) ENTER 11.29896425

So m∠B ≈ 11. By the Triangle Angle-Sum Theorem, m∠C ≈ 180 - (36 + 11) or 133.

Exercises

Solve each △ABC. Round measures of sides to the nearest tenth

and measures of angles to the nearest degree.

1. a = 9, b = 14, c = 12

2. m∠C = 80, c = 9, m∠A = 40

3. m∠B = 45, m∠C = 56, a = 2

4. a = 5.7, b = 6, c = 5

5. a = 11, b = 15, c = 21

8-6

Example

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Geometric Vector Operations A vector is a directed segment representing a quantity that has both magnitude, or length, and direction. For example, the speed and direction of an

airplane can be represented by a vector. In symbols, a vector is written as ! AB , where A is the

initial point and B is the endpoint, or as ! v . The sum of two vectors is called the resultant.

Subtracting a vector is equivalent to adding its opposite. The resultant of two vectors can be found using the parallelogram method or the triangle method.

Copy the vectors to find a -

b . a#

b#

Method 1: Use the parallelogram method.

Copy ! a and -

! b with the same initial

point.

a

-b

Complete the parallelogram.

a

-b

Draw the diagonal of the

parallelogram from the initial point.

a

a

-b

-b

Method 2: Use the triangle method.

Copy ! a .

a

Place the initial point of - ! b at the

terminal point of ! a .

a

-b

Draw the vector from the initial

point of ! a to the terminal point

of - ! b .

a

-b

a - b

Exercises

Copy the vectors. Then find each sum or difference.

1. ! c +

! d 2. w -

! z

3. ! a -

! b 4.

! r +

! t

##

Example

c# d#

w#

z#

r#

t#

a#

b#


Vectors

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Vectors

8-7

Vectors on the Coordinate Plane A vector in standard position has its initial point at (0, 0) and can be represented by the ordered

pair for point B. The vector at the right can be expressed as v = ⟨5, 3⟩.

You can use the Distance Formula to find the magnitude | AB | of a vector. You can describe the direction of a vector

by measuring the angle that the vector forms with the positive x-axis or with any other horizontal line.

Find the magnitude and direction of a = ⟨3, 5⟩.

Find the magnitude.

a = √ &&&&&&&&& (x

2 - x

1)2 + (y

2 - y

1)2

= √ &&&&&&&& (3 - 0)2 + (5 - 0)2

= √ && 34 or about 5.8

To find the direction, use the tangent ratio.

tan θ = 5 −

3 The tangent ratio is opposite over adjacent.

m∠θ ≈ 59.0 Use a calculator.

The magnitude of the vector is about 5.8 units and its direction is 59°.

Exercises

Find the magnitude and direction of each vector.

1. b = ⟨-5, 2⟩ 2. c = ⟨-2, 1⟩

3. d = ⟨3, 4⟩ 4. m = ⟨5, -1⟩

5. r = ⟨-3, -4⟩ 6. v = ⟨-4, 1⟩

x

y

(3, 5)

x

y

(5, 3)

(0, 0)

Example

Distance Formula

(x1, y

1) = (0, 0) and (x

2, y

2) = (3, 5)

Simplify.

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Skills Practice

Vectors

8-7

Use a ruler and a protractor to draw each vector. Include a scale on each

diagram.

1. a = 20 meters per second 2. b = 10 pound of force at

60° west of south 135° to the horizontal

Copy the vectors to find each sum or difference.

3. a + z 4. t -

r

Write the component form of each vector.

5. 6.


7. m = ⟨2, 12⟩ 8. p = ⟨3, 10⟩

9. k = ⟨-8, -3⟩ 10. f = ⟨-5, 11⟩

Find each of the following for a = ⟨2, 4⟩,

b = ⟨3, -3⟩ , and

c = ⟨4, -1⟩. Check your

answers graphically.

x

y

O

B(–2, 2)

D(3, –3)

x

y

O

C(1, –1)

E(4, 3)

a

z

r

t

N

E

S

W

1 cm : 5 m

60°a

1 in : 10 lb

135°b

Overmatter

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Practice

Vectors

Use a ruler and a protractor to draw each vector. Include a scale on each

diagram.

1. v = 12 Newtons of force at 2.

w = 15 miles per hour

40° to the horizontal 70° east of north


3. p +

r 4.

a -

b

5. Write the component form of AB .


6. t = ⟨6, 11⟩ 7.

g = ⟨9, -7⟩

Find each of the following for a = ⟨-1.5, 4⟩,

b = ⟨7, 3⟩, and

c = ⟨1, -2⟩. Check your

answers graphically.

8. 2 a +

b 9. 2

c -

b

10. AVIATION A jet begins a flight along a path due north at 300 miles per hour. A wind

is blowing due west at 30 miles per hour.

a. Find the resultant velocity of the plane.

b. Find the resultant direction of the plane.

8-7

x

y

O

K(–2, 4)

L(3, –4)

p%r%a% b%

1 cm : 4 N 40°

V

N

ES

W

1 in : 10 mi

70°

w

b 2a

x

y

−4

4

8

12

4−4 8 12

2a + b b

y

−4

−8

4

8

x

−8−12 4

2c - b

2c

Lesso

n 8

-7

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Vectors

8-7

1. WIND The vector v represents the speed

and direction that the wind is blowing.

Suddenly the wind picks up and doubles

its speed, but the direction does not

change. Write an expression for a vector

that describes the new wind velocity in

terms of v.

2. SWIMMING Jan is swimming in a

triathlon event. When the ocean water

is still, her velocity can be represented

by the vector ⟨2, 1⟩ miles per hour.

During the competition, there was a

fierce current represented by the

vector ⟨–1, –1⟩ miles per hour. What

vector represents Jan’s velocity during

the race?

3. POLYGONS Draw a regular polygon

around the origin. For each side of the

polygon, associate a vector whose

magnitude is the length of the

corresponding side and whose direction

points in the clockwise motion around

the origin. What vector represents the

sum of all these vectors? Explain.

4. BASEBALL Rick is in the middle of a

baseball game. His teammate throws

him the ball, but throws it far in front of

him. He has to run as fast as he can to

catch it. As he runs, he knows that as

soon as he catches it, he has to throw it

as hard as he can to the teammate at

home plate. He has no time to stop. In

the figure, x is the vector that represents

the velocity of the ball after Rick throws

it and v represents Rick’s velocity

because he is running. Assume that Rick

can throw just as hard when running as

he can when standing still.

Homeplate

Rick

x

v$

$

$X

$V

a. What vector would represent the

velocity of the ball if Rick threw it the

same way but he was standing still?

b. The angle between x and

v is 89°. By

running, did it help Rick get the ball

to home plate faster than he would

have normally been able to if he were

standing still?

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Enrichment8-7

Dot Product

The dot product of two vectors represents how much the vectors point in the direction of each other. If

v is a vector represented by ⟨a, b⟩ and

u is a vector represented by ⟨c, d⟩, the

formula to find the dot product is:

v ·

u = ac + bd

Look at the following example:

Graph the vectors and find the dot product of v and

u if

v = ⟨3, -1⟩ and

u = ⟨2, 5⟩.

v ·

u = (3)(2) + (-1)(5) or 1

Graph the vectors and find the dot products.

1. v = ⟨2, 1⟩ and

u = ⟨-4, 2⟩ 2.

v = ⟨3, -2⟩ and

u = ⟨1, 4⟩

The dot product is The dot product is

3. v = ⟨0, 3⟩ and

u = ⟨2, 4⟩ 4.

v = ⟨-1, 4⟩ and

u = ⟨-4, 2⟩

The dot product is The dot product is

5. Notice the angle formed by the two vectors and the corresponding dot product. Is there any relationship between the type of angle between the two vectors and the sign of the dot product? Make a conjecture.

y

xO

y

xO

y

xO

y

xO

y

xO

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Assessment


8

7.

Read each question. Then fill in the correct answer.

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing

each number or symbol in a box. Then fill in the corresponding circle for that

number or symbol.

Record your answers for Question 15 on the back of this paper.

Multiple Choice

Extended Response

Short Response/Gridded Response

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

7. (grid in)

8.

9.

10. (grid in)

11.

12.

13.

14.

10.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

Student Recording SheetUse this recording sheet with pages 610–611 of the Student Edition.

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8 Rubric for Scoring Extended-Response

General Scoring Guidelines

• If a student gives only a correct numerical answer to a problem but does not show how

he or she arrived at the answer, the student will be awarded only 1 credit. All extended-

response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all

parts of the question. For example, if a question has three parts, the correct response to

one or two parts of the question that required work to be shown is not considered a fully

correct response.

• Students who use trial and error to solve a problem must show their method. Merely

showing that the answer checks or is correct is not considered a complete response for

full credit.

Exercise 15 Rubric

Score Specific Criteria

4The values of x = 10, y = 16, and z = 12.8 are found using appropriate proportions based upon students knowledge of right triangle geometric mean theorems.

3A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0An incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given.

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Assessment

SCORE

NAME DATE PERIOD

Chapter 8 Quiz 2(Lessons 8-3 and 8-4)

8 SCORE


8 Chapter 8 Quiz 1(Lessons 8-1 and 8-2)

For Questions 1 and 2, find x.

1. 2.

For Questions 3 and 4, find x to the nearest tenth.

3. 4.

5. A rectangle has a diagonal 20 inches long that forms angles of

60° and 30° with the sides. Find the perimeter of the rectangle.

6. Find sin 52°. Round to the nearest ten-thousandth.

7. If cos A = 0.8945, find m∠A to the nearest degree.

8. The distance along a hill is 24 feet. If the land slopes uphill at

an angle of 8°, find the vertical distance from the top to the

bottom of the hill. Round to the nearest tenth.

1. Find the geometric mean between 12 and 16.

For Questions 2 and 3, find x and y.

2.

5 12

xy

3.

8

9

x

y

4. Find x. 4

11

x

5. The measures of the sides of a triangle are 19, 15, and 13. Use

the Pythagorean Theorem to classify the triangle as acute,

obtuse, or right.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

9. Use a calculator to find the

measure of ∠T to the

nearest tenth.

10. Use a calculator to

find the measure

of ∠T to the nearest

tenth.

17

31°

x

13

11

x°

45°

6x

60° 30°

6x

24

9

7121√3

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SCORE

SCORE


Chapter 8 Quiz 3(Lessons 8-5 and 8-6)

Chapter 8 Quiz 4(Lesson 8-7)

1. Name the angle of elevation in the figure.

2. Find x to the nearest tenth.

3. Solve △ABC. Round angle measures to the nearest degree and side measures to the nearest tenth.

4. Solve △RST. Round your answers to the nearest degree.

5. A squirrel, 37 feet up in a tree, sees a dog 29 feet from the base of the tree. Find the measure of the angle of depression to the nearest degree.

22°76°

10x

P

S

Q

R


1. " RT : R(14, 6) and T(−5, −10)

2. " PQ : P(20, −16) and Q(−9, −4)


3. " c + " d 4.

" k -

" m

5. Lidia is rollerblading south at a velocity of 9 miles per hour. The wind is blowing 3 miles per hour in the opposite direction. What is Lidia’s resultant velocity and direction?

1.

2.

3.

4.

5.

8

48 61

76T S

R

8

49°

11 18

A C

B

d

c

m%k%

1.

2.

3.

4.

5.

Assessment

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Chapter 8 Mid-Chapter Test(Lessons 8-1 through 8-4)

Part II

For Questions 6–8, find x and y.

6. 7.

8.

9. The measures of the sides of a triangle are 56, 90, and 106.

Use the Pythagorean Theorem to classify the triangle as acute,

obtuse, or right.

10. Guy wires 80 feet long support a 65-foot tall telephone pole. To

the nearest degree, what angle will the wires make with the

ground?

6.

7.

8.

9.

10.

1.

2.

3.

4.

5.


A 3 √ " 7 B 16 C 8 D 2

2. Find x.

F 6 √ " 6 H 6 √ " 55

G √ " 2

− 5

J √ " 23

− 5

3. Find sin C.

A √ " 2 C √ " 23

− √ " 2

B √ " 2 −

5 D

√ " 23 −

5


F 14 H 18.4

G 21.1 J 32.2

5. Find y to the nearest degree.

A 145 C 45

B 60 D 3527

19

y°

28

49°x

5√2

√"23B C

A

9

24

24

x

8

3

6 y

x

30°

60°

45°

8√3

y

x

60° 45°

20 y

x

Part I Write the letter for the correct answer in the blank at the right of each question.

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SCORE


Chapter 8 Vocabulary Test

Choose the correct word to complete each sentence.

1. The square root of the product of two numbers is the (geometric mean, or Pythagorean triple) of the numbers.

2. A group of three whole numbers that satisfy the equation a2 + b2 = c2, where c is the greatest number, is called a (trigonometric ratio, or Pythagorean triple).

Write whether each sentence is true or false. If false,

replace the underlined word or number to make a true

sentence.

3. The ratio of the lengths of any two sides of a right triangle is called a geometric mean.

4. An angle between the line of sight and the horizontal when anobserver looks upward is called a angle of elevation.

Choose from the terms above to complete each sentence.

5. An angle between the line of sight and the horizontal when an observer looks downward is called a(n) ? .

6. The word ? is derived from the Greek term for triangle and measure.

7. In a right triangle, the ? of an angle can be found by dividing the length of the opposite leg by the length of the triangle’s hypotenuse.

8. In a right triangle, the ? of an angle can be found by dividing the length of the opposite leg by the length of the adjacent leg.

Define each term in your own words.

9. component form

10. Pythagorean Theorem

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

angle of depression

angle of elevation

component form

cosine

direction

geometric mean

inverse cosine

inverse sine

inverse tangent

Law of Cosines

Law of Sines

magnitude

Pythagorean Triple

resultant

sine

standard position

tangent

trigonometric ratio

trigonometry

vector

8

Assessment

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Chapter 8 Test, Form 1

Write the letter for the correct answer in the blank at the right of each question.


A 100 B 50 C 12.5 D 10

2. Find x in △ABC.

F 8 H √ # 20

G 10 J 64

3. Find x in △PQR.

A 13 C 16

B 15 D √ # 60

4. Find x in △STU.

F 2 H √ # 32

G 8 J √ ## 514

5. Which set of measures could represent the lengths of the sides of a right

triangle?

A 2, 3, 4 C 8, 10, 12

B 7, 11, 14 D 9, 12, 15

6. Find x in △DEF.

F 6 H 6 √ # 3

G 6 √ # 2 J 12

7. Find y in △XYZ.

A 7.5 √ # 3 C 15

B 15 √ # 3 D 30

8. The length of the sides of a square is 10 meters. Find the length of the

diagonals of the square.

F 10 m H 10 √ # 3 m

G 10 √ # 2 m J 20 m

9. Find x in △HJK.

A 5 √ # 2 C 10

B 5 √ # 3 D 15

10. Find x in △ABC.

F 25 H 25 √ # 3

G 25 √ # 2 J 100

D E

F

6

6x

S T

U

15

17x

P Q

R

12

5x

A B

C 4

16x

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

8

A C

B

60° 30°

50

x

H J

K

60°

30°

5

x

X Y

Z

15√245°

45°

y

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Chapter 8 Test, Form 1 (continued)


A 7.3 C 18.4

B 17.3 D 47.1

12. Find the measure of the angle of elevation of the Sun when a pole 25 feet tall casts a shadow 42 feet long.

F 30.8° G 36.5° H 53.5° J 59.2°

13. Which is the angle of depression in the figure at the right?

A ∠AOT C ∠TOB

B ∠AOB D ∠BTO

14. Find y in △XYZ if m∠Y = 36, m∠X = 49, and x = 12. Round to the nearest hundreth.

F 0.04 G 9.35 H 14.80 J 15.41

15. To find the distance between two points, A and B, on opposite sides of a river, a surveyor measures the distance from A to C as 200 feet, m∠A = 72, and m∠B = 37. Find the distance from A to B. Round your answer to the nearest tenth.

A 77.4 ft B 201.2 ft C 250.4 ft D 314.2 ft

16. In △ABC, m∠A = 40, m∠C = 115, and b = 8. Find a to the nearest tenth.

F 12.2 G 11.3 H 5.7 J 5.3

17. Find the length of the third side of a triangular garden if two sides measure 8 feet and 12 feet and the included angle measures 50.

A 7.8 ft B 9.2 ft C 14.4 ft D 146.3 ft

18. In △DEF, d = 20, e = 25, and f = 30. Find m∠F to the nearest degree.

F 83° G 76° H 56° J 47°

19. Find the component form of # AB with A(2, 3) and B(−4, 6).

A ⟨−2, 9⟩ B ⟨2, −9⟩ C ⟨−6, 3⟩ D ⟨6, −3⟩

20. Find the magnitude of # AB with A(3, 4) and B(−1, 7).

F ⟨4, −3⟩ G 5 H √ ** 13 J 25

Bonus In △ABC, a = 50, b = 48, and c = 40. Find m∠ A to the nearest degree.

A B

C

AO

B T

67° 20

x

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

8

Assessment

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Chapter 8 Test, Form 2A



A 5 C √ " 19

B 9.5 D 2 √ " 21

2. In △PQR, RS = 4 and QS = 6. Find PS.

F 2 H √ " 10

G 5 J 2 √ " 6

3. Find x.

A 3 √ " 2 C 4.5

B √ " 14 D 3

4. Find y.

F 12 H 9

G 11 J 2

5. Find the length of the hypotenuse of a right triangle with legs that measure

5 and 7.

A 12 C √ " 35

B √ " 24 D √ " 74

6. Find x.

F 3 H 4 √ " 3

G 4 J 2 √ " 5


triangle?

A 9, 40, 41 C 7, 8, 15

B 8, 30, 31 D √ " 2 , √ " 3 , √ " 6

8. Find c.

F 7 H 7 √ " 3

G 7 √ " 2 J 14

9. Find the perimeter of a square if the length of its diagonal is 12 inches.

Round to the nearest tenth.

A 8.5 in. C 48 in.

B 33.9 in. D 67.9 in.

10. Find x.

F 4 H 4 √ " 3

G 4 √ " 2 J 8 √ " 3 x

88

60°

45°

7c

x66

8

3

y 6

2 7

x

P Q

S

R4

6

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

8

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Chapter 8 Test, Form 2A (continued)


A 5.8 C 8.1

B 5.9 D 17.3

12. Find x to the nearest degree.

F 56 H 34

G 45 J 29

13. If a 20-foot ladder makes a 65° angle with the ground, how many feet up a wall will it reach? Round your answer to the nearest tenth.

A 8.5 ft B 10 ft C 18.1 ft D 42.9 ft

14. A ship’s sonar finds that the angle of depression to a wreck on the bottom of the ocean is 12.5°. If a point on the ocean floor is 60 meters directly below the ship, how many meters is it from that point on the ocean floor to the wreck? Round your answer to the nearest tenth.

F 277.2 m G 270.6 m H 61.5 m J 13.3 m

15. Find the angle of elevation of the sun if a building 100 feet tall casts a shadow 150 feet long. Round to the nearest degree.

A 60° B 48° C 42° D 34°

16. When the Sun’s angle of elevation is 73°, a tree tilted at an angle of 5° from the vertical casts a 20-foot shadow on the ground. Find the length of the tree to the nearest tenth of a foot.

F 6.3 ft H 51.1 ft

G 19.2 ft J 219.4 ft

17. In △CDE, m∠C = 52, m∠D = 17, and e = 28.6. Find c to the nearest tenth.

A 77.1 B 49.1 C 24.1 D 18.4

18. In △PQR, p = 56, r = 17, and m∠Q = 110. Find q to the nearest tenth.

F 4076.2 G 63.8 H 52.6 J 3.1

19. Find the component form of $ CD with C(5,-7) and D(-3, 9).

A ⟨-2, 2⟩ B ⟨2, 2⟩ C ⟨8, -16⟩ D ⟨-8, 16⟩

20. A pilot is flying due east at a speed of 300 miles per hour and wind is blowing due north at 50 miles per hour. What is the magnitude of the resultant velocity of the plane?

F 300 mph G 350 mph H about 304 mph J 2500 mph

Bonus From a window 20 feet above the ground, the angle of elevation to the top of another building is 35°. The distance between the buildings is 52 feet. Find the height of the building to the nearest tenth of a foot.

73°

5°

20-foot shadow

✹

95

x°

10

36°

x

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

8

Assessment

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Chapter 8 Test, Form 2B



A 3 √ " 11 C 10

B 2 √ " 5 D 2

2. In △PQR, RS = 5 and QS = 8. Find PS.

F 3 H √ " 13

G 6.5 J 2 √ " 10

3. Find x.

A 5.5 C √ " 24

B √ " 11 D √ " 33

4. Find y.

F 4 H 8

G 5 J 9

5. Find the length of the hypotenuse of a right triangle whose legs measure

6 and 5.

A 11 C √ " 30

B √ " 11 D √ " 61

6. Find x.

F √ " 39 H 5 √ " 3

G 6 J 5


triangle?

A 3 −

4 , 1,

5 −

4 C 7, 17, 24

B √ " 3 , √ " 5 , √ " 15 D 8, 15, 16

8. Find c.

F 18 H 9 √ " 2

G 9 √ " 3 J 9

9. Find the perimeter of a square if the length of its diagonal is 16 millimeters.

Round to the nearest tenth.

A 11.3 mm C 90.5 mm

B 45.3 mm D 128.0 mm

10. Find x.

F 6 H 6 √ " 3

G 6 √ " 2 J 12 √ " 3

x1212

60°

45°

9c

x

8

8

10

y

4

6

x

83

P Q

SR 5

8

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

8

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Chapter 8 Test, Form 2B (continued)

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

11. Find x.

A 8.0 C 10.4

B 8.9 D 10.8


F 57 H 33

G 55 J 29

13. If a 24-foot ladder makes a 58° angle with the ground, how many feet up a wall will it reach? Round your answer to the nearest tenth.

A 38.4 ft B 20.8 ft C 20.4 ft D 12.7 ft

14. A ship’s sonar finds that the angle of depression to a wreck on the bottom of the ocean is 13.2°. If a point on the ocean floor is 75 meters directly below the ship, how many meters is it from that point on the ocean floor to the wreck? Round to the nearest tenth.

F 328.4 m G 319.8 m H 77.0 m J 17.6 m

15. Find the angle of elevation of the sun if a building 125 feet tall casts a shadow 196 feet long. Round to the nearest degree.

A 64° B 50° C 40° D 33°

16. When the sun’s angle of elevation is 76°, a tree tilted at an angle of 4° from the vertical casts an 18-foot shadow on the ground. Find the length of the tree to the nearest tenth of a foot.

F 250.4 ft H 17.7 ft

G 56.5 ft J 4.6 ft

17. In △ABC, m∠A = 46, m∠B = 105, and c = 19.8. Find a to the nearest tenth.

A 29.4 B 28.5 C 15.7 D 14.7

18. In △LMN, ℓ = 42, m = 61, and m∠N = 108. Find n to the nearest tenth.

F 7068.4 G 84.1 H 79.2 J 24.7

19. Find the component from of $ EF with E(-11,-3) and F(7,-4)

A ⟨-18, 1⟩ B ⟨18, -1⟩ C ⟨-4, -7⟩ D ⟨4, 7⟩

20. An eagle is traveling along a path due east at a rate of 50 miles per hour and the wind is blowing due north at 15 miles per hour. What is the magnitude of the resultant velocity of the eagle?

F 35 mph G 50 mph H 65 mph J about 52.2 mph

Bonus From a window 24 feet above the ground, the angle of elevation to the top of another building is 38°. The distance between the buildings is 63 feet. Find the height of the building to the nearest tenth of a foot.

76°

4°

18-foot shadow

✹

x°

116

x

12

42°

B:

8

Assessment

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SCORE


Chapter 8 Test, Form 2C


For Questions 2–5, find x.

2. 3.

4. 5.

6. Find x.

7. In parallelogram ABCD, AD = 4

and m∠D = 60. Find AF.

8. Find x and y.


10. An A-frame house is 40 feet high and

30 feet wide. Find the measure of the

angle that the roof makes with the floor.

Round to the nearest degree.

11. A 30-foot tree casts a 12-foot shadow. Find the measure of the

angle of elevation of the Sun to the nearest degree.

9.2

18°

x

A D

C

F

B

22

x

x2020

20

60

20x

2 12

x

6

4x

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

8

4√3

60°

30°

x

y

30 ft

x°

40 ft

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Chapter 8 Test, Form 2C (continued)

12. A boat is 1000 meters from a cliff. If the angle of depression from the top of the cliff to the boat is 15°, how tall is the cliff? Round your answer to the nearest tenth.

13. A plane flying at an altitude of 10,000 feet begins descending when the end of the runway is 50,000 feet from a point on the ground directly below the plane. Find the measure of the angle of descent (depression) to the nearest degree.



16. A tree grew at a 3° slant from the vertical. At a point 50 feet from the tree, the angle of elevation to the top of the tree is 17°. Find the height of the tree to the nearest tenth of a foot.


18. In △XYZ, m∠X = 152, y = 15, and z = 19. Find x to the nearest tenth.

19. Find the magnitude and direction of the vector % AZ : A(8, 8)

and Z(1, −3).

20. Copy the vectors to find % s +

% t .

Bonus Find x.

11

75

x°

17° 93°

50 ft

x

x7

23°

146°

52°

26

37°x

1000 m

8

√6

5x

t*s*

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Assessment

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SCORE


Chapter 8 Test, Form 2D

1. Find the geometric mean between 3 √ " 6 and 5 √ " 6 .

For Questions 2–5, find x.

2. 3.

4. 5.

6. Find x.

7. In parallelogram ABCD, AD = 14

and m∠D = 60. Find AF.

8. Find x and y.


10. An A-frame house is 45 feet high and

32 feet wide. Find the measure of the

angle that the roof makes with the

floor. Round to the nearest degree.

11. A 38-foot tree casts a 16-foot shadow. Find the measure of the

angle of elevation of the sun to the nearest degree.

16°

8.3

x

A D

C

F

B

30

x

x2424

24

80

30x

103

x

12

8

x

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

8

8√3

30°

60°

x

y

32 ft

x

45 ft

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Chapter 8 Test, Form 2D (continued)

12. A boat is 2000 meters from a cliff. If the angle of depression from the top of the cliff to the boat is 10°, how tall is the cliff? Round your answer to the nearest tenth.

13. A plane flying at an altitude of 10,000 feet begins descending when the end of the runway is 60,000 feet from a point on the ground directly below the plane. Find the measure of the angle of descent (depression) to the nearest degree.



16. A tree grew at a 3° slant from the vertical. At a point 60 feet from the tree, the angle of elevation to the top of the tree is 14°. Find the height of the tree to the nearest tenth of a foot.


18. In △XYZ, m∠X = 156, y = 18, and z = 21. Find x to the nearest tenth.

19. Find the magnitude and direction of the vector $ PQ : P(−2, 4)

and Q(−5, −6).

20. Copy the vectors to find $ d -

$ f .

Bonus Find x.

8 9

16

x°

14° 93°

60 ft

x

38°

17

6

x°

68°

42°

23x

2000 m

8

2√15

16

x

d* f*

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Assessment

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SCORE


Chapter 8 Test, Form 3

1. Find the geometric mean between 2−9

and 3−9

.

2. Find x in △PQR.

3. Find x in △XYZ.

4. If the length of one leg of a right triangle is three times the length of the other and the hypotenuse is 20, find the length of the shorter leg.

5. Find the measure of the altitude drawn to the hypotenuse of a right triangle with legs that measure 3 and 4.

6. Find x.

7. Richmond is 200 kilometers due east of Teratown and Hamilton is 150 kilometers directly north of Teratown. Find the shortest distance in kilometers between Hamilton and Richmond.

8. The measures of the sides of a triangle are 48, 55, and 73. Use the Pythagorean Theorem to classify the triangle as acute, obtuse, or right.

9. Find the exact perimeter of this square.

10. Find the exact perimeter of rectangle ABCD.

11. Find x and y.

12. △ABC is a 30°-60°-90° triangle with right angle A and

with −−AC as the longer leg. If A(-4, -2) and B(-4, 6), find the

coordinates of C.

15

30°

60°

xy

12

60°

A B

CD

17

8

9 3.5

x

65

2x

Q

P

R

S

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

8

x + 4 xX

WZ

Y

21√$

36√

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Chapter 8 Test, Form 3 (continued)

13. If −−

AB ‖ −−−

CD , find x and the exact length of

−−− CD .

14. The angle of elevation from a point on the street to the top of a building is 29°. The angle of elevation from another point on the street, 50 feet farther away from the building, to the top of the building is 25°. To the nearest foot, how tall is the building?

15. The angle of depression from the top of a flagpole on top of a lighthouse to a boat on the ocean is 2.9°. The angle of depression from the bottom of the flagpole to the boat is 2.6°. If the boat is 400 feet away from shore and the lighthouse is right on the edge of the shore, how tall is the flagpole? Round your answer to the nearest foot.

16. In △JKL, m∠J = 26.8, m∠K = 19, and k = 17. Find ℓ to the nearest tenth.

17. Don hit a golf ball from the tee toward the hole which is 250 yards away. However, due to the wind, his drive was 5° off course. If the angle between the segment from the hole to the tee and the segment from the hole to the ball measures 97°, how far did Don drive the ball? Round to the nearest tenth of a yard.

18. In △HJK, h = 7, j = 12.3, and k = 7.9. Find m∠K. Round your answer to the nearest degree.

19. Find the magnitude and direction of the vector (WZ : W(15, 25)

and Z(10, −6).

20. Copy the vectors to find (a +

(b and

(a -

(b .

Bonus A 50-foot vertical pole that stands on a hillside. The slope of the hill side makes an angle of 10° with the horizontal. Two guy wires extend from the top of the pole to points on the hill 60 feet uphill and downhill from its base. Find the length of each guy wire to the nearest tenth of a foot.

12

10

60°45°x

A B

C D

8

13.

14.

15.

16.

17.

18.

19.

20.

B:

a*b*

Assessment

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Chapter 8 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to

each problem. Be sure to include all relevant drawings and justify

your answers. You may show your solution in more than one way or

investigate beyond the requirements of the problem.

1. If the geometric mean between 10 and x is 6, what is x? Show how you

obtained your answer.

2.

a. Max used the following equations to find x in △PQR. Is Max correct?

Explain.

x = √ ## 8 · 2

x = √ ## 16

x = 4

b. If ∠PRQ is a right angle, what is the measure of −−−

PS ?

c. Is △PRS a 45°-45°-90° triangle? Explain.

3. To solve for x in a triangle, when would you use sin and when would you

use sin-1? Give an example for each type of situation.

4. Draw a diagram showing the angles of elevation and depression and label

each. How are the measures of these angles related?

5. Draw an obtuse triangle and label the vertices, the measures of two

angles, and the length of one side. Explain how to solve the triangle.

6. Irina is solving △ABC. She plans to first use the Law of Sines to find two

of the angles. Is Irina’s plan a good one? Explain.

4

15

12

A C

B

8 2

60°

x

P Q

R

S

8

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SCORE


Standardized Test Practice (Chapters 1–8)

1. If −− TA bisects ∠YTB,

−− TC bisects ∠BTZ, m∠YTA = 4y + 6, and m∠BTC = 7y - 4, find m∠CTZ. (Lesson 1-4)

A 52 B 38 C 25 D 8

2. Which statement is always true? (Lesson 2-5)

F If right triangle QPR has sides q, p, and r, where r is the hypotenuse,

then r2 = p2 + q2.

G If −− EF ‖

−− HJ , then EF = HJ.

H If −− KL and

−− VT are cut by a transversal, then −− KL ‖ −− VT .

J If −−− DR and

−−− RH are congruent, then R bisects −−− DH .

3. The equation for $ %& PT is y - 2 = 8(x + 3). Determine an equation for a

line perpendicular to $ %& PT . (Lesson 3-4)

A y = 1 − 8 x - 7 B y = 8x - 13 C y = - 1 −

8 x + 2 D y = -8x

4. Angle Y in △XYZ measures 90°. −− XY and

−− YZ each measure 16 meters. Classify △XYZ. (Lesson 4-1)

F acute and isosceles H right and scalene

G equiangular and equilateral J right and isosceles

5. Two sides of a triangle measure 4 inches and 9 inches. Determine which cannot be the perimeter of the triangle. (Lesson 5-3)

A 19 in. B 21 in. C 23 in. D 26 in.

6. △ABC ∼ △STR, so AB −

CA = ? . (Lesson 6-2)

F AB

− BC

G ST

− RS

H TR

− RS

J RS

− ST

7. The Petronas Towers in Kuala Lumpur, Malaysia, are 452 meters tall. A woman who is 1.75 meters tall stands 120 meters from the base of one tower. Find the angle of elevation between the woman’s hat and the top of the tower. Round to the nearest tenth. (Lesson 8-5)

A 14.8° B 34.9° C 55° D 75.1°

8. Which equation can be used to find x? (Lesson 8-4)

F x = y sin 73° H x = y −

cos 73°

G x = y cos 73° J x = y −

sin 73°

9. Which inequality describes the possible values

of x? (Lesson 5-6)

A x < 4 C x > 2B x < 2 D x > 4

85°117°

1212

14

5x - 6

73°

x

y

Y ZT

A CB

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

8

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

Assessment

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15.

16.

Standardized Test Practice (continued)

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

10. Ashley wants to make a poster for her campaign from a photograph. She uses a photocopier to enlarge the 4 inch by 6 inch photograph. What are the dimensions of the poster if she increases the size of the photograph by a scale factor of 5? (Lesson 7-7)

F 2000 inches by 3000 inches H 9 inches by 11 inches

G 0.8 inches by 1.2 inches J 20 inches by 30 inches

11. Find x. (Lesson 6-2)

A -3 C 5

B 3 D 9

12. Trapezoid ABCD has vertices A(1, 6), B(-2, 6), C(-10, -10), and

D(20, -10). Find the measure of ABCD’s midsegment to the nearest tenth. (Lesson 6-6)

F 3 G 5.3 H 7.2 J 16.5

For Questions 13 and 14, use the

figure to the right.

13. Find QP to the nearest tenth. (Lesson 8-2)

A 7.5 B 12 C 18.3 D 19.6

14. Find LM. (Lesson 8-3)

F 5 G 5 √ " 3 H 9 J 10 √ " 3

60°30°

10

9

P

M

L

Q

R

5

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. Find x so that ℓ ‖ m. (Lesson 3-5)

16. Find c to the nearest tenth. (Lesson 8-6)

(134 - 4x)°(6x + 17)°

mℓ

8

Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column box and then

shading in the appropriate circle that corresponds to that entry.

26.1°

27.7°

126.2°

19

33

A

B

C

c

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Standardized Test Practice (continued)

For Questions 17 and 18, complete

the following proof. (Lesson 2-7)

Given: −−

JK " −−−

LM

−−−

HJ " −−

KL

Prove: −−−

HK " −−−

KM

Proof:

Statements Reasons

1. −−

JK " −−−

LM , −−−

HJ " −−

KL 1. Given

2. JK = LM, HJ = KL 2. (Question 17)

3. (Question 18) 3. Segment Addition Post.

4. HJ + JK = KL + LM 4. Add. Prop. of Equality

5. HK = KM 5. Substitution Prop.

6. −−−

HK " −−−

KM 6. Def. of " segments

For Questions 19 and 20, use the figure

at the right.

19. Find the measure of the numbered angles if m∠ABC = 57 and m∠BCE = 98. (Lesson 4-2)

20. If −−−

BD is a midsegment of △ABC, AD = 2x - 6, and DC = 22.5 - 4x, find AC. (Lesson 5-2)

21. If △DEF " △HJK, m∠D = 26, m∠J = 3x + 5, and m∠F = 92, find x. (Lesson 4-3)

22. Use the Exterior Angle Inequality Theorem to list all of the angles with measures that are less than m∠1. (Lesson 5-3)

23. a. Determine whether △EFH ∼ △JGH. Justify your answer. (Lesson 7-3)

b. If G is the midpoint of −−−

FH , find x. (Lesson 8-3)

c. Use the value of x you found in part b to find the scale factor of △EFH to △JGH. (Lesson 7-3)

1 2 4

3 5

6 7

8A DC

B

E

CB

A

D

25°1

2

34

5

41°

H

J

K L M

17.

18.

19.

20.

21.

22.

23a.

23b.

23c.

Part 3: Short Response

Instructions: Write your answer in the space provided.

60°18

30°E JH

F

Gx

8

Answers

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Chapter 8 A1 Glencoe Geometry

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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s of

the l

egs.

D

4

. If

an

y t

rian

gle

has

sid

es

wit

h l

en

gth

s 3,

4,

an

d 5

, th

en

th

at

tria

ngle

is

a r

igh

t tr

ian

gle

.A

5

. If

th

e t

wo a

cute

an

gle

s of

a r

igh

t tr

ian

gle

are

45°,

then

th

e l

en

gth

of

the h

yp

ote

nu

se i

s √

"

2 t

imes

the

len

gth

of

eit

her

leg.

A

6

. In

an

y t

rian

gle

wh

ose

an

gle

measu

res

are

30°,

60°,

an

d 9

0°,

th

e h

yp

ote

nu

se i

s √

"

3 t

imes

as

lon

g a

s th

e

short

er

leg.

D

7

. T

he s

ine r

ati

o o

f an

an

gle

of

a r

igh

t tr

ian

gle

is

equ

al

to t

he l

en

gth

of

the a

dja

cen

t si

de d

ivid

ed

by t

he

len

gth

of

the h

yp

ote

nu

se.

D

8

. T

he t

an

gen

t of

an

an

gle

of

a r

igh

t tr

ian

gle

wh

ose

si

des

have l

en

gth

s 3,

4,

an

d 5

wil

l be s

mall

er

than

th

e

tan

gen

t of

an

an

gle

of

a r

igh

t tr

ian

gle

wh

ose

sid

es

have l

en

gth

s 6,

8,

an

d 1

0.

D

9

. T

rigon

om

etr

ic r

ati

os

can

be u

sed

to s

olv

e p

roble

ms

involv

ing a

ngle

s of

ele

vati

on

an

d a

ngle

s of

dep

ress

ion

.A

10

. T

he L

aw

of

Sin

es

can

on

ly b

e u

sed

in

rig

ht

tria

ngle

s.D

Ste

p 2

Lesson 8-1


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

5

Gle

nc

oe

Ge

om

etr

y

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Geo

metr

ic M

ean

Ge

om

etr

ic M

ea

n

Th

e g

eo

me

tric

me

an

betw

een

tw

o n

um

bers

is

the p

osi

tive s

qu

are

ro

ot

of

their

pro

du

ct.

For

two p

osi

tive n

um

bers

a a

nd

b,

the g

eom

etr

ic m

ean

of

a a

nd

b i

s

the p

osi

tive n

um

ber

x i

n t

he p

rop

ort

ion

a

−

x =

x −

b .

Cro

ss m

ult

iply

ing g

ives

x2 =

ab,

so x

= √

""

ab .

F

ind

th

e g

eo

me

tric

me

an

be

twe

en

ea

ch

pa

ir o

f n

um

be

rs.

a.

12

an

d 3

x =

√ "

"

ab

Definitio

n o

f geom

etr

ic m

ean

=

√ "

""

12 . 3

a

= 1

2 a

nd b

= 3

=

√ "

""

""

(2 .

2 .

3)

. 3

Facto

r.

=

6

Sim

plif

y.

Th

e g

eom

etr

ic m

ean

betw

een

12 a

nd

3 i

s 6.

b.

8 a

nd

4

x =

√ "

"

ab

Definitio

n o

f geom

etr

ic m

ean

=

√ "

"

8 . 4

a

= 8

and b

= 4

=

√ "

""

"

(2 .

4)

. 4

Facto

r.

=

√ "

""

16 .

2

Associa

tive P

ropert

y

=

4 √

"

2

Sim

plif

y.

Th

e g

eom

etr

ic m

ean

betw

een

8 a

nd

4

is 4

√ "

2 o

r abou

t 5.7

.

Exerc

ises

Fin

d t

he

ge

om

etr

ic m

ea

n b

etw

ee

n e

ach

pa

ir o

f n

um

be

rs.

1. 4 a

nd

4

2. 4 a

nd

6

3. 6 a

nd

9

4. 1

−

2 a

nd

2

5. 12 a

nd

20

6. 4 a

nd

25

7. 16 a

nd

30

8. 10 a

nd

100

9. 1

−

2 a

nd

1

−

4

10. 17 a

nd

3

11. 4 a

nd

16

12. 3 a

nd

24

8-1 Exam

ple

4

√ "

"

24 o

r 2 √

"

6 ≈

4.9

√ "

"

240 o

r 4 √

""

15 ≈

15.5

√ "

""

1000 o

r 10 √

""

10 ≈

31.6

10

√"

"54

or

3√

" 6≈

7.3

1

√ "

"

480 o

r 4 √

""

30 ≈

21.9

√ "

"

51 ≈

7.1

8

√ "

"

72 o

r 6 √

"

2 ≈

8.5

√ "

1

−

8 o

r √

"

2

−

4 ≈

0.4

Answers (Anticipation Guide and Lesson 8-1)

Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

6

Gle

nc

oe

Ge

om

etr

y

z

y

x

15

R QP

S25

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Geo

metr

ic M

ean

Ge

om

etr

ic M

ea

ns

in R

igh

t T

ria

ng

les

In t

he d

iagra

m,

△ A

BC

∼ △

AD

B ∼

△ B

DC

. A

n a

ltit

ud

e t

o t

he h

yp

ote

nu

se o

f a r

igh

t tr

ian

gle

form

s tw

o r

igh

t tr

ian

gle

s. T

he t

wo t

rian

gle

s are

sim

ilar

an

d

each

is

sim

ilar

to t

he o

rigin

al

tria

ngle

.

U

se

rig

ht

△ A

BC

wit

h

−−

BD

⊥ −

−

AC

. D

escrib

e t

wo

ge

om

etr

ic

me

an

s.

a.

△ A

DB

∼ △

BD

C s

o A

D

−

BD

= B

D

−

CD

.

In △

AB

C,

the a

ltit

ud

e i

s th

e g

eom

etr

ic

mean

betw

een

th

e t

wo s

egm

en

ts o

f th

e

hyp

ote

nu

se.

b.

△ A

BC

∼ △

AD

B a

nd △

AB

C ∼

△ B

DC

,

so

AC

−

AB

= A

B

−

AD

an

d A

C

−

BC

= B

C

−

DC

.

In △

AB

C,

each

leg i

s th

e g

eom

etr

ic

mean

betw

een

th

e h

yp

ote

nu

se a

nd

th

e

segm

en

t of

the h

yp

ote

nu

se a

dja

cen

t to

that

leg.

F

ind

x,

y,

an

d z

.

15 =

√ &

&&

&

RP

' S

P

Geom

etr

ic M

ean (

Leg)

Theore

m

15 =

√ &

&

25x

RP

= 2

5 a

nd S

P =

x

225 =

25x

Square

each s

ide.

9 =

x

Div

ide e

ach s

ide b

y 2

5.

Th

en

y =

RP

– S

P

=

25 –

9

=

16

z =

√ &

&&

&

RS

' R

P

Geom

etr

ic M

ean (

Leg)

Theore

m

=

√ &

&&

16 '

25

RS

= 1

6 a

nd R

P =

25

=

√ &

&

400

Multip

ly.

=

20

Sim

plif

y.

Exerc

ises

Fin

d x

, y,

an

d z

to

th

e n

ea

re

st

ten

th.

1.

x

13

2.

z

xy

5

2

3.

zx

y 81

x =

√ %

3 ≈

1.7

x =

√ %

%

10 ≈

3.2

; x =

3;

y =

√ %

%

14 ≈

3.7

; y =

√ %

%

72 o

r 6 √

%

2 ≈

8.5

;

z =

√ %

%

35 ≈

5.9

z =

√ %

8 o

r 2 √

%

2 ≈

2.8

4.

xy

1

√3

√"12

5.

x

zy

2

2

6.

xz

y

62

x =

2;

x =

2;

x =

√ %

%

12 o

r 2 √

%

3 ≈

3.5

;

y =

3

y =

√ %

8 o

r 2 √

%

2 ≈

2.8

; y =

√ %

8 o

r 2 √

%

2 ≈

2.8

;

z =

√ %

8 o

r 2 √

%

2 ≈

2.8

z =

√ %

%

24 o

r 2 √

%

6 ≈

4.9

8-1

CDB

A

Exam

ple

1Exam

ple

2

Lesson 8-1


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

7

Gle

nc

oe

Ge

om

etr

y

Sk

ills

Pra

ctic

e

Geo

metr

ic M

ean

Fin

d t

he

ge

om

etr

ic m

ea

n b

etw

ee

n e

ach

pa

ir o

f n

um

be

rs.

1. 2 a

nd

8

2. 9 a

nd

36

3. 4 a

nd

7

4. 5 a

nd

10

5. 28 a

nd

14

6. 7 a

nd

36

Writ

e a

sim

ila

rit

y s

tate

me

nt

ide

nti

fyin

g t

he

th

re

e s

imil

ar t

ria

ng

les i

n t

he

fig

ure

.

7.

C

D

B

A

8.

L

M N

P

9.

GEH

F

10.

RT

S U

Fin

d x

, y a

nd

z.

11.

39y

x

z

12.

10

4

y

x

z

13.

15

4

y

xz

14.

25y

x

z

8-1

4

18

√

%%

28 o

r 2 √

%

7 ≈

5.3

√ %

%

50 o

r 5 √

%

2 ≈

7.1

√ %

%

392 o

r 14 √

%

2 ≈

19.8

√

%%

252 o

r 6 √

%

7 ≈

15.9

△A

CB

∼ △

CD

B ∼

△A

DC

△M

NL ∼

△N

PL ∼

△M

PN

△E

GF ∼

△G

HF ∼

△E

HG

△R

ST ∼

△S

UT ∼

△R

US

6;

√ %

%

108 o

r 6 √

%

3 ≈

10.4

;

√ %

%

40 o

r 2 √

%%

10 ≈

6.3

; √

%%

56 o

r 2 √

%%

14 ≈

7.5

;

√ %

%

27 o

r 3 √

%

3 ≈

5.2

√

%%

140 o

r 2 √

%%

35 ≈

11.8

√%

%60

or

2√

%%

15

≈ 7

.7;

12.5

; √

%%

29

≈ 5

.4;

√%

%%

181.2

5≈

13.5

√ %

%

285 o

r 2 √

%%

19 ≈

16.9

;

√ %

%

76 ≈

8.7

Answers (Lesson 8-1)

Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

8

Gle

nc

oe

Ge

om

etr

y

Practi

ce

Geo

metr

ic M

ean

Fin

d t

he

ge

om

etr

ic m

ea

n b

etw

ee

n e

ach

pa

ir o

f n

um

be

rs.

1. 8 a

nd

12

2. 3 a

nd

15

3.

4

−

5 a

nd

2

Writ

e a

sim

ila

rit

y s

tate

me

nt

ide

nti

fyin

g t

he

th

re

e s

imil

ar t

ria

ng

les i

n t

he

fig

ure

.

4.

U

TA

V

5.

KLJ

M

Fin

d x

, y,

an

d z

.

6.

23

z

xy

8

7.

zx

y625

8.

x

y

2

3

z

9.

xy1

0z

20

10

. C

IVIL

A

n a

irp

ort

, a f

act

ory

, an

d a

sh

op

pin

g c

en

ter

are

at

the v

ert

ices

of

a r

igh

t tr

ian

gle

form

ed

by t

hre

e h

igh

ways.

Th

e a

irp

ort

an

d f

act

ory

are

6.0

mil

es

ap

art

. T

heir

dis

tan

ces

from

th

e s

hop

pin

g c

en

ter

are

3.6

mil

es

an

d 4

.8 m

iles,

resp

ect

ively

. A

serv

ice r

oad

wil

l be

con

stru

cted

fro

m t

he s

hop

pin

g c

en

ter

to t

he h

igh

way t

hat

con

nect

s th

e a

irp

ort

an

d

fact

ory

. W

hat

is t

he s

hort

est

poss

ible

len

gth

for

the s

erv

ice r

oad

? R

ou

nd

to t

he n

eare

st

hu

nd

red

th.

8-1

√ "

"

96 o

r 4 √

"

6 ≈

9.8

√ "

"

45 o

r 3 √

"

5 ≈

6.7

√ "

8

−

5 o

r 2

√ "

"

10

−

5

≈ 1

.3

△

VU

T ∼

△U

AT ∼

△V

AU

△

JLK

∼ △

LM

K ∼

△JM

L

x

= √

""

184 o

r 2 √

""

46 ≈

13.6

; x

= √

""

114 ≈

10.7

;

y=

√"

"248

or

2√

""

62

≈ 1

5.7

y=

√"

"150

or

5√

" 6≈

12.2

z

= √

""

713 ≈

26.7

z

= √

""

475 o

r 5 √

""

19 ≈

21.8

x

= 4

.5;

x

= 1

5;

x

= √

""

13 ≈

3.6

; 6.5

y =

5;

z =

√ "

"

300 o

r 10 √

"

3 ≈

17.3

2.8

8 m

i

Lesson 8-1


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

9

Gle

nc

oe

Ge

om

etr

y

1. S

QU

AR

ES

W

ilm

a h

as

a r

ect

an

gle

of

dim

en

sion

s ℓ b

y w

. S

he w

ou

ld l

ike t

o

rep

lace

it

wit

h a

squ

are

th

at

has

the

sam

e a

rea.

Wh

at

is t

he s

ide l

en

gth

of

the s

qu

are

wit

h t

he s

am

e a

rea a

s

Wil

ma’s

rect

an

gle

?

2. E

QU

ALIT

Y G

retc

hen

com

pu

ted

th

e

geom

etr

ic m

ean

of

two n

um

bers

. O

ne o

f

the n

um

bers

was

7 a

nd

th

e g

eom

etr

ic

mean

tu

rned

ou

t to

be 7

as

well

. W

hat

was

the o

ther

nu

mber?

3. V

IEW

ING

AN

GLE

A

ph

oto

gra

ph

er

wan

ts t

o t

ak

e a

pic

ture

of

a b

each

fro

nt.

His

cam

era

has

a v

iew

ing a

ngle

of

90°

an

d h

e w

an

ts t

o m

ak

e s

ure

tw

o p

alm

trees

loca

ted

at

poin

ts A

an

d B

in

th

e

figu

re a

re j

ust

in

sid

e t

he e

dges

of

the

ph

oto

gra

ph

.

Walkway

90 f

t4

0 f

tA

B

x

AB

x

He w

alk

s ou

t on

a w

alk

way t

hat

goes

over

the o

cean

to g

et

the s

hot.

If

his

cam

era

has

a v

iew

ing a

ngle

of

90°,

at

wh

at

dis

tan

ce d

ow

n t

he w

alk

way s

hou

ld

he s

top

to t

ak

e h

is p

hoto

gra

ph

?

4

. E

XH

IBIT

ION

S A

mu

seu

m h

as

a f

am

ou

s

statu

e o

n d

isp

lay.

Th

e c

ura

tor

pla

ces

the

statu

e i

n t

he c

orn

er

of

a r

ect

an

gu

lar

room

an

d b

uil

ds

a 1

5-f

oot-

lon

g r

ail

ing i

n

fron

t of

the s

tatu

e.

Use

th

e i

nfo

rmati

on

belo

w t

o f

ind

how

clo

se v

isit

ors

wil

l be

able

to g

et

to t

he s

tatu

e.

Sta

tue

12

ft

15 ft

9 f

tx

5

. C

LIF

FS

A

bri

dge c

on

nect

s to

a t

un

nel

as

show

n i

n t

he f

igu

re.

Th

e b

rid

ge i

s

180 f

eet

above t

he g

rou

nd

. A

t a d

ista

nce

of

235 f

eet

alo

ng t

he b

rid

ge o

ut

of

the

tun

nel,

th

e a

ngle

to t

he b

ase

an

d

sum

mit

of

the c

liff

is

a r

igh

t an

gle

.

Clif

f

23

5 f

t

180 ft

dx

h

a.

Wh

at

is t

he h

eig

ht

of

the c

liff

? R

ou

nd

to t

he n

eare

st w

hole

nu

mber.

b.

How

hig

h i

s th

e c

liff

fro

m b

ase

to

sum

mit

? R

ou

nd

to t

he n

eare

st w

hole

nu

mber.

c.

Wh

at

is t

he v

alu

e o

f d

? R

ou

nd

to t

he

neare

st w

hole

nu

mber.

Wo

rd

Pro

ble

m P

racti

ce

Geo

metr

ic M

ean

8-1

7

.2 f

t

6

0 f

t

387 f

t

√ "

"

ℓw

, th

e g

eo

metr

ic m

ean

of

ℓ

an

d w

7

4

87 f

t

3

07 f

t


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

10

G

len

co

e G

eo

me

try

Math

em

ati

cs a

nd

Mu

sic

Pyth

agora

s, a

Gre

ek

ph

iloso

ph

er

wh

o l

ived

du

rin

g t

he s

ixth

cen

tury

B.C

.,

beli

eved

th

at

all

natu

re,

beau

ty,

an

d h

arm

on

y c

ou

ld b

e e

xp

ress

ed

by w

hole

-n

um

ber

rela

tion

ship

s. M

ost

peop

le r

em

em

ber

Pyth

agora

s fo

r h

is t

each

ings

abou

t ri

gh

t tr

ian

gle

s. (

Th

e s

um

of

the s

qu

are

s of

the l

egs

equ

als

th

e s

qu

are

of

the h

yp

ote

nu

se.)

Bu

t P

yth

agora

s als

o d

isco

vere

d r

ela

tion

ship

s betw

een

th

e

mu

sica

l n

ote

s of

a s

cale

. T

hese

rela

tion

ship

s ca

n b

e e

xp

ress

ed

as

rati

os.

C

D

E

F

G

A

B

C

′1

−

1

8

−

9

4

−

5

3

−

4

2

−

3

3

−

5

8

−

15

1

−

2

Wh

en

you

pla

y a

str

inged

in

stru

men

t,

Th

e C

str

ing c

an

be u

sed

you

pro

du

ce d

iffe

ren

t n

ote

s by p

laci

ng

to p

rod

uce

F b

y p

laci

ng

you

r fi

nger

on

dif

fere

nt

pla

ces

on

a s

trin

g.

a f

inger

3

−

4 o

f th

e w

ay

Th

is i

s th

e r

esu

lt o

f ch

an

gin

g t

he l

en

gth

alo

ng t

he s

trin

g.

of

the v

ibra

tin

g p

art

of

the s

trin

g.

Su

pp

ose

a C

str

ing

ha

s a

le

ng

th o

f 1

6 i

nch

es.

Writ

e a

nd

so

lve

pro

po

rti

on

s t

o d

ete

rm

ine

wh

at

len

gth

of

str

ing

wo

uld

ha

ve

to

vib

ra

te t

o p

ro

du

ce

th

e r

em

ain

ing

no

tes o

f th

e s

ca

le.

1

. D

2. E

3. F

4

. G

5. A

6. B

7. C

′

8

. C

om

ple

te t

o s

how

th

e d

ista

nce

betw

een

fin

ger

posi

tion

s on

th

e 1

6-i

nch

C

str

ing f

or

each

note

. F

or

exam

ple

, C

(16)

- D

(14 2

−

9 ) =

1 7

−

9 .

C

D

E

F

G

A

B

C

′

3 4of

C s

trin

g

En

rich

men

t

1 7 − 9

in

.

8-1

3 1 − 5

in

.4 i

n.

5 1 − 3

in

.6 2 − 5

in

.7 7

−

15 in

.8 i

n.

10 2

−

3 in

.8

8

−

15 in

.

12 i

n.

8 i

n.

14 2 − 9

in

.12 4 − 5

in

.

9 3

−

5 in

.

Lesson 8-2


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

11

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Th

e P

yth

ag

ore

an

Th

eo

rem

an

d I

ts C

on

vers

e

Th

e P

yth

ag

ore

an

Th

eo

rem

In

a r

igh

t tr

ian

gle

, th

e s

um

of

the

squ

are

s of

the l

en

gth

s of

the l

egs

equ

als

th

e s

qu

are

of

the l

en

gth

of

the h

yp

ote

nu

se.

If t

he t

hre

e w

hole

nu

mbers

a,

b,

an

d c

sati

sfy t

he e

qu

ati

on

a

2 +

b2 =

c2,

then

th

e n

um

bers

a,

b,

an

d c

form

a

Py

tha

go

re

an

trip

le.

ca

bA

CB

a.

Fin

d a

.

a

12

13

ACB

a

2 +

b2 =

c2

Pyth

agore

an T

heore

m

a

2 +

12

2 =

13

2

b =

12,

c =

13

a

2 +

144 =

169

Sim

plif

y.

a

2 =

25

Subtr

act.

a

= 5

T

ake t

he p

ositiv

e s

quare

root

of

each s

ide.

b.

Fin

d c

. c

30

20

ACB

a

2 +

b2 =

c2

Pyth

agore

an T

heore

m

20

2 +

30

2 =

c2

a =

20,

b =

30

400 +

900 =

c2

Sim

plif

y.

1300 =

c2

Add.

√

$$

1300 =

c

Take t

he p

ositiv

e s

quare

root

of

each s

ide.

36.1

≈ c

U

se a

calc

ula

tor.

Fin

d x

.

1.

x

33

2.

x 15

9

3.

x65

25

4.

x

5 9

4 9

5.

x

33

16

6.

x

11

28

Use

a P

yth

ag

ore

an

Trip

le t

o f

ind

x.

7

. 8

17

x

8

. 45

24

x

9

. 28

96

x

8-2 Exam

ple

Exerc

ises

△ A

BC

is a

rig

ht

tria

ngle

.

so a

2 +

b2 =

c2.

1 − 3√

##

#1345

≈ 3

6.7

√#

#663

≈ 2

5.7

√ #

#

18 o

r 3

√ #

2 ≈

4.2

12

60

15

51

100

Answers (Lesson 8-1 and Lesson 8-2)

Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

12

G

len

co

e G

eo

me

try

c

ab

A

C

B

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Th

e P

yth

ag

ore

an

Th

eo

rem

an

d I

ts C

on

vers

e

Co

nv

ers

e o

f th

e P

yth

ag

ore

an

Th

eo

rem

If

th

e s

um

of

the s

qu

are

s

of

the l

en

gth

s of

the t

wo s

hort

er

sid

es

of

a t

rian

gle

equ

als

th

e s

qu

are

of

the l

en

gth

s of

the l

on

gest

sid

e,

then

th

e t

rian

gle

is

a r

igh

t tr

ian

gle

.

You

can

als

o u

se t

he l

en

gth

s of

sid

es

to c

lass

ify a

tri

an

gle

. If a

2 +

b2 =

c2,

then

if

a2

+ b

2 =

c2 t

hen

△A

BC

is

a r

igh

t tr

ian

gle

. △

AB

C is a

rig

ht

tria

ngle

.

if

a2

+ b

2 >

c2 t

hen

△A

BC

is

acu

te.

if

a2

+ b

2 <

c2 t

hen

△A

BC

is

obtu

se.

D

ete

rm

ine

wh

eth

er △

PQ

R i

s a

rig

ht

tria

ng

le.

a

2 +

b2 "

c2

Com

pare

c2

and a

2 +

b2

10

2 +

(10 √

$

3 )2

" 2

02

a =

10,

b =

10 √

$

3 ,

c =

20

100 +

300 "

400

Sim

plif

y.

400 =

400

%

Add.

Sin

ce c

2 =

an

d a

2 +

b2,

the t

rian

gle

is

a r

igh

t tr

ian

gle

.

Exerc

ises

De

term

ine

wh

eth

er e

ach

se

t o

f m

ea

su

re

s c

an

be

th

e m

ea

su

re

s o

f th

e s

ide

s o

f a

tria

ng

le.

If s

o,

cla

ssif

y t

he

tria

ng

le a

s a

cu

te,

ob

tuse,

or r

igh

t. J

usti

fy y

ou

r a

nsw

er.

1. 30,

40,

50

2. 20,

30,

40

3. 18,

24,

30

4. 6,

8,

9

5.

6,

12,

18

6. 10,

15,

20

7. √

$

5 ,

√ $

$

12 ,

√ $

$

13

8.

2,

√ $

8 ,

√ $

$

12

9. 9,

40,

41

20

10

10√3

8-2 Exam

ple

y

es,

rig

ht;

yes,

ob

tuse;

yes,

rig

ht;

50

2 =

30

2 +

40

2

40

2 >

20

2 +

30

2

30

2 =

24

2 +

18

2

y

es,

acu

te;

n

o;

6 +

12 =

18

yes,

ob

tuse;

92 <

62 +

82

20

2 >

10

2 +

15

2

y

es,

acu

te;

yes,

rig

ht;

y

es,

rig

ht;

(

√ #

#

13 ) 2

< ( √

#

5 ) 2

+ (

√ #

#

12 ) 2

( √ #

#

12 ) 2

= ( √

#

8 ) 2

+ 2

2

41

2 =

40

2 +

92


NA

ME

DA

TE

PE

RIO

D

Lesson 8-2

Ch

ap

ter

8

13

G

len

co

e G

eo

me

try

Sk

ills

Pra

ctic

e

Th

e P

yth

ag

ore

an

Th

eo

rem

an

d I

ts C

on

vers

e

Fin

d x

.

1.

x 12

9

2.

x

1213

3.

x

12

32

4.

x12.5

25

5.

x9

9

8

6.

x31

14

Use

a P

yth

ag

ore

an

Trip

le t

o f

ind

x.

7.

5

12x

8.

8

10

x

9.

20

12

x

10

.

65

25

x

11

. 1

2.

5048

x

De

term

ine

wh

eth

er e

ach

se

t o

f n

um

be

rs c

an

be

me

asu

re

of

the

sid

es o

f a

tria

ng

le.

If s

o,

cla

ssif

y t

he

tria

ng

le a

s a

cu

te,

ob

tuse,

or r

igh

t. J

usti

fy y

ou

r a

nsw

er.

13

. 7,

24,

25

14

. 8,

14,

20

15

. 12.5

, 13,

26

16

. 3 √

"

2 ,

√ "

7 ,

4

17

. 20,

21,

29

18

. 32,

35,

70

8-2

40

24

x

1

5

5

√

##

#

1168 ≈

34.2

√#

##

468.7

5≈

21.7

√#

#65

≈ 8

.1√

##

#1157

≈ 3

4.0

Y

es,

rig

ht

tria

ng

le

Yes,

ob

tuse t

rian

gle

N

o,

12.5

+ 1

3 <

26

7 2 +

24 2 =

25 2

8 2 +

14 2 <

20 2

1

3

6

16

6

0

32

14

Y

es,

acu

te t

rian

gle

Yes,

rig

ht

tria

ng

le

No

, 32 +

35 <

70

( 3 √

#

2 ) 2

+ √

#

7 2 >

42

20 2 +

21 2

= 2

9 2


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

14

G

len

co

e G

eo

me

try

Practi

ce

Th

e P

yth

ag

ore

an

Th

eo

rem

an

d I

ts C

on

vers

e

Fin

d x

.

1.

x

13

23

2.

x

34

21

3.

x

26

26

18

4.

x

34

22

5.

x

16

14

6.

x

24

24

42

Use

a P

yth

ag

ore

an

Trip

le t

o f

ind

x.

7.

36

27

x

8.

120

136

x

9.

65

39

x

10.

42

15

0

x

De

term

ine

wh

eth

er e

ach

se

t o

f n

um

be

rs c

an

be

me

asu

re

of

the

sid

es o

f a

tria

ng

le.

If s

o,

cla

ssif

y t

he

tria

ng

le a

s a

cu

te,

ob

tuse,

or r

igh

t. J

usti

fy y

ou

r a

nsw

er.

11. 10,

11,

20

12. 12,

14,

49

13. 5 √

"

2 ,

10,

11

14. 21.5

, 24,

55.5

15. 30,

40,

50

16. 65,

72,

97

17

. C

ON

ST

RU

CT

ION

T

he b

ott

om

en

d o

f a r

am

p a

t a w

are

hou

se i

s 10 f

eet

from

th

e b

ase

of

the m

ain

dock

an

d i

s 11 f

eet

lon

g.

How

h

igh

is

the d

ock

?

11 f

t?

dock

ram

p

10 f

t

8-2

√ "

"

698 ≈

26.4

√ "

"

715 ≈

26.7

√ "

"

595 ≈

24.4

√ "

""

1640 ≈

40.5

√ "

"

60 ≈

7.7

√ "

"

135 ≈

11.6

ab

ou

t 4.6

ft

hig

h

4

5

5

2

y

es,

ob

tuse t

rian

gle

; n

o,

12 +

14

< 4

9;

yes,

acu

te t

rian

gle

;

1

0 2 +

11 2 <

20 2

(

5 √

"

2 ) 2

+ 1

0 2 >

11 2

n

o,

21.5

+ 2

4 <

55.5

y

es,

rig

ht

tria

ng

le;

yes,

rig

ht

tria

ng

le;

3

0 2 +

40 2 =

50 2

65 2 +

72 2 =

97 2

64

144


NA

ME

DA

TE

PE

RIO

D

Lesson 8-2

Ch

ap

ter

8

15

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

Th

e P

yth

ag

ore

an

Th

eo

rem

an

d I

ts C

on

vers

e

1

. S

IDE

WA

LK

S C

on

stru

ctio

n w

ork

ers

are

bu

ild

ing a

marb

le s

idew

alk

aro

un

d a

p

ark

th

at

is s

hap

ed

lik

e a

rig

ht

tria

ngle

. E

ach

marb

le s

lab a

dd

s 2 f

eet

to t

he

len

gth

of

the s

idew

alk

. T

he w

ork

ers

fi

nd

th

at

exact

ly 1

071 a

nd

1840 s

labs

are

requ

ired

to m

ak

e t

he s

idew

alk

s alo

ng t

he s

hort

sid

es

of

the p

ark

. H

ow

m

an

y s

labs

are

requ

ired

to m

ak

e t

he

sid

ew

alk

th

at

run

s alo

ng t

he l

on

g s

ide

of

the p

ark

?

2

. R

IGH

T A

NG

LE

S C

lyd

e m

ak

es

a t

rian

gle

u

sin

g t

hre

e s

tick

s of

len

gth

s 20 i

nch

es,

21 i

nch

es,

an

d 2

8 i

nch

es.

Is

the t

rian

gle

a r

igh

t tr

ian

gle

? E

xp

lain

.

3

. T

ET

HE

RS

T

o h

elp

su

pp

ort

a f

lag p

ole

, a

50-f

oot-

lon

g t

eth

er

is t

ied

to t

he p

ole

at

a p

oin

t 40 f

eet

above t

he g

rou

nd

. T

he

teth

er

is p

ull

ed

tau

t an

d t

ied

to a

n

an

chor

in t

he g

rou

nd

. H

ow

far

aw

ay

from

th

e b

ase

of

the p

ole

is

the a

nch

or?

4

. FLIG

HT

A

n a

irp

lan

e l

an

ds

at

an

air

port

60 m

iles

east

an

d 2

5 m

iles

nort

h o

f w

here

it

took

off

.

60 m

i

25

mi

H

ow

far

ap

art

are

th

e t

wo a

irp

ort

s?

5

. P

YT

HA

GO

RE

AN

TR

IPLE

S M

s. J

on

es

ass

ign

ed

her

fift

h-p

eri

od

geom

etr

y c

lass

th

e f

oll

ow

ing p

roble

m.

Let

m a

nd

n b

e t

wo p

osi

tive i

nte

gers

w

ith

m >

n.

Let

a =

m2 –

n2,

b =

2m

n,

an

d c

= m

2 +

n2.

a.

Sh

ow

th

at

there

is

a r

igh

t tr

ian

gle

w

ith

sid

e l

en

gth

s a

, b,

an

d c

.

b.

Com

ple

te t

he f

oll

ow

ing t

able

.

mn

ab

c

21

34

5

31

86

10

32

512

13

41

15

817

42

12

16

20

43

724

25

51

24

10

26

c.

Fin

d a

Pyth

agore

an

tri

ple

th

at

corr

esp

on

ds

to a

rig

ht

tria

ngle

wit

h

a h

yp

ote

nu

se 2

52 =

625 u

nit

s lo

ng.

(Hin

t: U

se t

he t

able

you

com

ple

ted

fo

r E

xerc

ise b

to f

ind

tw

o p

osi

tive

inte

gers

m a

nd

n w

ith

m >

n a

nd

m

2 +

n2 =

625.)

8-2

6

5 m

i

2

129

3

0 f

t

n

o;

20 2 +

21 2 ≠

28 2

S

am

ple

an

sw

er:

a 2 +

b 2 =

( m

2 –

n 2 ) 2

+ (

2m

n) 2

= m

4 -

2 m

2 n

2 +

n 4 +

4 m

2 n

2 =

m 4 +

2 m

2 n

2 +

n 4

an

d c

2 =

( m

2 +

n 2 ) 2

= m

4 +

2 m

2 n

2 +

n 4 .

Th

is m

ean

s t

hat

a 2 +

b 2 =

c 2 ,

so

th

at

a,

b,

an

d c

do

fo

rm t

he s

ides o

f a r

igh

t

tria

ng

le b

y t

he c

on

vers

e o

f th

e

Pyth

ag

ore

an

Th

eo

rem

.

S

am

ple

an

sw

er:

Take m

= 2

4

an

d n

= 7

to

get

a =

527,

b =

336,

an

d c

= 6

25.


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

16

G

len

co

e G

eo

me

try

En

rich

men

t

Co

nvers

e o

f a R

igh

t Tri

an

gle

Th

eo

rem

You

have l

earn

ed

th

at

the m

easu

re o

f th

e a

ltit

ud

e f

rom

th

e v

ert

ex o

f th

e r

igh

t an

gle

of

a r

igh

t tr

ian

gle

to i

ts h

yp

ote

nu

se i

s th

e g

eom

etr

ic

mean

betw

een

th

e m

easu

res

of

the t

wo s

egm

en

ts o

f th

e h

yp

ote

nu

se.

Is t

he c

on

vers

e o

f th

is t

heore

m t

rue?

In o

rder

to f

ind

ou

t, i

t w

ill

help

to

rew

rite

th

e o

rigin

al

theore

m i

n i

f-th

en

form

as

foll

ow

s.

If △

AB

Q i

s a r

igh

t tr

ian

gle

wit

h r

igh

t an

gle

at

Q,

then

QP

is

the g

eom

etr

ic m

ean

betw

een

AP

an

d P

B,

wh

ere

P

is b

etw

een

A a

nd

B a

nd

−−

−

QP

is

perp

en

dic

ula

r to

−−

AB

.

1

. W

rite

th

e c

on

vers

e o

f th

e i

f-th

en

form

of

the t

heore

m.

2

. Is

th

e c

on

vers

e o

f th

e o

rigin

al

theore

m t

rue?

Refe

r to

th

e f

igu

re a

t th

e r

igh

t to

exp

lain

you

r an

swer.

Y

ou

may f

ind

it

inte

rest

ing t

o e

xam

ine t

he o

ther

theore

ms

in

Ch

ap

ter

8 t

o s

ee w

heth

er

their

con

vers

es

are

tru

e o

r fa

lse.

You

wil

l n

eed

to r

est

ate

th

e t

heore

ms

care

full

y i

n o

rder

to w

rite

th

eir

co

nvers

es.

Q

BP

A

Q

BP

A

8-2

If

QP

is t

he g

eo

metr

ic m

ean

betw

een

AP

an

d

PB

, w

here

P i

s b

etw

een

A a

nd

B a

nd

−−

QP

⊥ −

−

AB

,

then

△A

BQ

is a

rig

ht

tria

ng

le w

ith

rig

ht

an

gle

at

Q.

Y

es;

(PQ

) 2 =

(A

P)(

PB

) im

plies t

hat

PQ

−

AP

= P

B

−

PQ

.

S

ince b

oth

∠A

PQ

an

d ∠

QP

B a

re r

igh

t an

gle

s,

they a

re c

on

gru

en

t. T

here

fore

△A

PQ

∼ △

QP

B b

y S

AS

sim

ilari

ty.

So

∠A

& ∠

PQ

B a

nd

∠A

QP

& ∠

B.

Bu

t th

e a

cu

te

an

gle

s o

f △

AQ

P a

re c

om

ple

men

tary

an

d

m∠

AQ

B =

m∠

AQ

P +

m∠

PQ

B.

Hen

ce

m∠

AQ

B =

90 a

nd

△A

QB

is a

rig

ht

tria

ng

le

wit

h r

igh

t an

gle

at

Q.


NA

ME

DA

TE

PE

RIO

D

Lesson 8-2

Ch

ap

ter

8

17

G

len

co

e G

eo

me

try

Sp

read

sheet

Act

ivit

y

Pyth

ag

ore

an

Tri

ple

s

You

can

use

a s

pre

ad

sheet

to d

ete

rmin

e w

heth

er

thre

e w

hole

nu

mbers

fo

rm a

Pyth

agore

an

tri

ple

.

U

se

a s

pre

ad

sh

ee

t to

de

term

ine

wh

eth

er t

he

nu

mb

ers 1

2,

16

, a

nd

20

form

a P

yth

ag

ore

an

trip

le.

Ste

p 1

In

cell

A1,

en

ter

12.

In c

ell

B1,

en

ter

16 a

nd

in

cell

C1,

en

ter

20.

Th

e lo

nges

t si

de

shou

ld b

e en

tere

d i

n c

olu

mn

C.

Ste

p 2

In

cell

D1,

en

ter

an

equ

als

sig

n f

oll

ow

ed

by

IF(A

1^

2+

B1^

2=

C1^

2,“

YE

S”,

“NO

”).

Th

is w

ill

retu

rn “

YE

S”

if t

he s

et

of

nu

mbers

is

a

Pyth

agore

an

tri

ple

an

d w

ill

retu

rn “

NO

” if

it

is n

ot.

Th

e n

um

bers

12,

16,

an

d 2

0 f

orm

a P

yth

agore

an

tri

ple

.

U

se

a s

pre

ad

sh

ee

t to

de

term

ine

wh

eth

er t

he

nu

mb

ers 3

, 6

, a

nd

12

form

a P

yth

ag

ore

an

trip

le.

Ste

p 1

In

cell

A2,

en

ter

3,

in c

ell

B2,

en

ter

6,

an

d i

n c

ell

C2,

en

ter

12.

Ste

p 2

C

lick

on

th

e b

ott

om

rig

ht

corn

er

of

cell

D1 a

nd

dra

g i

t to

D2.

Th

is

wil

l d

ete

rmin

e w

heth

er

or

not

the s

et

of

nu

mbers

is

a P

yth

agore

an

tr

iple

.

Th

e n

um

bers

3,

6,

an

d 1

2 d

o n

ot

form

a P

yth

agore

an

tri

ple

.

Exerc

ises

Use

a s

pre

ad

sh

ee

t to

de

term

ine

wh

eth

er e

ach

se

t o

f n

um

be

rs

form

s a

Py

tha

go

re

an

trip

le.

1. 14,

48,

50

2

. 16,

30,

34

3

. 5,

5,

9

4

. 4,

5,

7

5. 18,

24,

30

6. 10,

24,

26

7

. 25,

60,

65

8

. 2,

4,

5

9. 19,

21,

22

10

. 18,

80,

82

1

1. 5,

12,

13

1

2. 20,

48,

52

8-2

Exam

ple

1

Exam

ple

2

y

es

yes

no

n

o

yes

yes

y

es

no

n

o

y

es

yes

yes


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

18

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Sp

ecia

l R

igh

t Tri

an

gle

s

Pro

pe

rtie

s o

f 4

5°-

45

°-9

0°

Tri

an

gle

s T

he s

ides

of

a 4

5°-

45°-

90°

righ

t tr

ian

gle

have a

sp

eci

al

rela

tion

ship

.

If

th

e l

eg

of

a 4

5°-

45

°-9

0°

rig

ht

tria

ng

le i

s x

un

its,

sh

ow

tha

t th

e h

yp

ote

nu

se

is x

√ "

2

un

its.

x√

x

x2

45

°

45

°

Usi

ng t

he P

yth

agore

an

Th

eore

m w

ith

a

= b

= x

, th

en

c2

= a

2 +

b2

c2

= x

2 +

x2

c2

=

2x

2

c

= √

##

2x

2

c

= x

√ #

2

In

a 4

5°-

45

°-9

0°

rig

ht

tria

ng

le t

he

hy

po

ten

use

is √

"

2 t

ime

s

the

le

g.

If t

he

hy

po

ten

use

is 6

un

its,

fin

d t

he

le

ng

th o

f e

ach

le

g.

Th

e h

yp

ote

nu

se i

s √

#

2 t

imes

the l

eg,

so

div

ide t

he l

en

gth

of

the h

yp

ote

nu

se b

y √

#

2 .

a =

6

−

√ #

2

=

6

−

√ #

2 .

√ #

2

−

√ #

2

=

6 √

#

2

−

2

=

3 √

#

2 u

nit

s

Exerc

ises

Fin

d x

.

1.

x

8

45

°

45

°

2.

x

45

°3

√2

3.

45°

4

x

4.

xx

18

5.

45°

16

x

x

6.

45°

24

x

√2

7. If

a 4

5°-

45°-

90°

tria

ngle

has

a h

yp

ote

nu

se l

en

gth

of

12,

fin

d t

he l

eg l

en

gth

.

8

. D

ete

rmin

e t

he l

en

gth

of

the l

eg o

f 45°-

45°-

90°

tria

ngle

wit

h a

hyp

ote

nu

se l

en

gth

of

25 i

nch

es.

9. F

ind

th

e l

en

gth

of

the h

yp

ote

nu

se o

f a 4

5°-

45°-

90°

tria

ngle

wit

h a

leg l

en

gth

of

14 c

en

tim

ete

rs.

8-3

Exam

ple

1Exam

ple

2

9 √

"

2 ≈

12.7

8 √

"

2 ≈

11.3

24

8 √

"

2 ≈

11.3

3

4 √

"

2 ≈

5.7

6 √

"

2 ≈

8.5

25

√ "

2

−

2

in

. ≈

17.7

in

.

14 √

"

2 c

m ≈

19.8

cm

Lesson 8-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

19

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Sp

ecia

l R

igh

t Tri

an

gle

s

Pro

pe

rtie

s o

f 3

0°-

60

°-9

0°

Tri

an

gle

s T

he s

ides

of

a 3

0°-

60°-

90°

righ

t tr

ian

gle

als

o

have a

sp

eci

al

rela

tion

ship

.

In

a 3

0°-

60

°-9

0°

rig

ht

tria

ng

le t

he

hy

po

ten

use

is t

wic

e t

he

sh

orte

r l

eg

. S

ho

w t

ha

t th

e l

on

ge

r l

eg

is √

"

3 t

ime

s

the

sh

orte

r l

eg

.

△ M

NQ

is

a 3

0°-

60°-

90°

righ

t tr

ian

gle

, an

d t

he l

en

gth

of

the

hyp

ote

nu

se −

−−

MN

is

two t

imes

the l

en

gth

of

the s

hort

er

sid

e −−

− N

Q .

Use

th

e P

yth

agore

an

Th

eore

m.

a2 =

(2

x)2

- x

2

a2 =

c2 -

b2

a2 =

4x

2 -

x2

M

ultip

ly.

a2 =

3x

2

Subtr

act.

a =

√ #

#

3x

2

Take t

he p

ositiv

e s

quare

root

of

each s

ide.

a =

x √

#

3

Sim

plif

y.

In

a 3

0°-

60

°-9

0°

rig

ht

tria

ng

le,

the

hy

po

ten

use

is 5

ce

nti

me

ters.

Fin

d t

he

le

ng

ths o

f th

e o

the

r t

wo

sid

es o

f th

e t

ria

ng

le.

If t

he h

yp

ote

nu

se o

f a 3

0°-

60°-

90°

righ

t tr

ian

gle

is

5 c

en

tim

ete

rs,

then

th

e l

en

gth

of

the

short

er

leg i

s on

e-h

alf

of

5,

or

2.5

cen

tim

ete

rs.

Th

e l

en

gth

of

the l

on

ger

leg i

s √

#

3 t

imes

the

len

gth

of

the s

hort

er

leg,

or

(2.5

)( √

#

3 )

cen

tim

ete

rs.

Exerc

ises

Fin

d x

an

d y

.

1.

x y30

°

60

°1 2

2.

x

y

60

°

8

3.

x

y11

30

°

4.

xy

30

°

9√

3

5.

xy

60

°

12

6.

60°

x20

y

7. A

n e

qu

ilate

ral

tria

ngle

has

an

alt

itu

de l

en

gth

of

36 f

eet.

Dete

rmin

e t

he l

en

gth

of

a s

ide

of

the t

rian

gle

.

8

. F

ind

th

e l

en

gth

of

the s

ide o

f an

equ

ilate

ral

tria

ngle

th

at

has

an

alt

itu

de l

en

gth

of

45 c

en

tim

ete

rs.

8-3

30°

60°

x2

xaExam

ple

1

Exam

ple

2

x =

1;

x =

8 √

"

3 ≈

13.9

;

x =

5.5

;

y =

0.5

√ "

3 ≈

0.9

y =

16

y =

5.5

√ "

3 ≈

9.5

x =

9;

x =

4 √

"

3 ≈

6.9

;

x =

10 √

"

3 ≈

17.3

;

y =

18

y =

8 √

"

3 ≈

13.9

y

= 1

0

24 √

"

3 f

eet

≈ 4

1.6

ft

30 √

"

3 c

m ≈

52 c

m


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

20

G

len

co

e G

eo

me

try

8-3

Sk

ills

Practi

ce

Sp

ecia

l R

igh

t Tri

an

gle

s

Fin

d x

.

1.

45°

25

x

2

.

45°

17

x

3

. 45°

48

x

4

.

45°

100

x

5

.

45°

100

x

6

.

45°88

x

7. D

ete

rmin

e t

he l

en

gth

of

the l

eg o

f 45°-

45°-

90°

tria

ngle

wit

h a

hypote

nu

se l

en

gth

of

26.

8. F

ind

th

e l

en

gth

of

the h

yp

ote

nu

se o

f a 4

5°-

45°-

90°

tria

ngle

wit

h a

leg l

en

gth

of

50 c

en

tim

ete

rs.

Fin

d x

an

d y

.

9.

30°

x11

y

1

0.

60°

x

8

y

√3

1

1.

30°

x

5

y

√3

12

.

60°

x

30

y

1

3.

30°

x

21

y√3

1

4.

60°

x

52

y√3

15. A

n e

qu

ilate

ral

tria

ngle

has

an

alt

itu

de l

en

gth

of

27 f

eet.

Dete

rmin

e t

he l

en

gth

of

a s

ide

of

the t

rian

gle

.

16

. F

ind

th

e l

en

gth

of

the s

ide o

f an

equ

ilate

ral

tria

ngle

th

at

has

an

alt

itu

de l

en

gth

of

11 √

"

3 f

eet.

2

5 √

"

2

8.5

√ "

2 o

r 1

7 √

"

2

−

2

4

8 √

"

2

2

2;

11 √

"

3

12;

4 √

"

3

15;

10 √

"

3

5

0 √

"

2

100 √

"

2

44 √

"

2

13 √

"

2

5

0 √

"

2 c

m

1

5;

15 √

"

3

42;

21

78;

26 √

"

3

1

8 √

"

3

2

2

Lesson 8-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

21

G

len

co

e G

eo

me

try

Fin

d x

.

1.

45°

14

x

2.

45°

45

x

3.

45°

22

x

4.

45°

210

x

5.

45°

88

x

6.

45°

x5

√2

Fin

d x

an

d y

.

7.

30°

x9

y

8.

60°

x

4

y

√3

9.

30°

x

20

y

1

0.

60°

x

98

y

11

. D

ete

rmin

e t

he l

en

gth

of

the l

eg o

f 45°-

45°-

90°

tria

ngle

wit

h a

hyp

ote

nu

se l

en

gth

of

38.

12. F

ind

th

e l

en

gth

of

the h

yp

ote

nu

se o

f a 4

5°-

45°-

90°

tria

ngle

wit

h a

leg l

en

gth

of

77 c

en

tim

ete

rs.

13. A

n e

qu

ilate

ral

tria

ngle

has

an

alt

itu

de l

en

gth

of

33 f

eet.

Dete

rmin

e t

he l

en

gth

of

a s

ide

of

the t

rian

gle

.

14

. B

OT

AN

ICA

L G

AR

DE

NS

O

ne o

f th

e d

isp

lays

at

a b

ota

nic

al

gard

en

is a

n h

erb

gard

en

pla

nte

d i

n t

he s

hap

e o

f a s

qu

are

. T

he s

qu

are

measu

res

6 y

ard

s on

each

sid

e.

Vis

itors

can

vie

w t

he h

erb

s fr

om

a

dia

gon

al

path

way t

hro

ugh

th

e g

ard

en

. H

ow

lon

g i

s th

e p

ath

way?

6 yd

6 yd

6 yd

6 yd

Practi

ce

Sp

ecia

l R

igh

t Tri

an

gle

s

8-3

1

4 √

"

2

22.5

√ "

2 o

r 4

5 √

"

2

−

2

2

2 √

"

2

1

05 √

"

2

88 √

"

2

10

x

= 1

8;

x

= 6

;

y

= 9

√ "

3

y =

2 √

"

3

x

= 2

0 √

"

3 ;

y

= 4

0

1

9 √

"

2

7

7 √

"

2 c

m

2

2 √

"

3 f

t

6

√ "

2 y

d o

r ab

ou

t 8.4

9 y

d

x =

49;

y =

49 √

"

3


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

22

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

Sp

ecia

l R

igh

t Tri

an

gle

s

1

. O

RIG

AM

I A

squ

are

pie

ce o

f p

ap

er

150 m

illi

mete

rs o

n a

sid

e i

s fo

lded

in

half

alo

ng a

dia

gon

al.

Th

e r

esu

lt i

s a

45°-

45°-

90°

tria

ngle

. W

hat

is t

he l

en

gth

of

the h

yp

ote

nu

se o

f th

is t

rian

gle

?

2

. E

SC

ALA

TO

RS

A

40-f

oot-

lon

g e

scala

tor

rise

s fr

om

th

e f

irst

flo

or

to t

he s

eco

nd

floor

of

a s

hop

pin

g m

all

. T

he e

scala

tor

mak

es

a 3

0°

an

gle

wit

h t

he h

ori

zon

tal.

40 f

t

30

°

? f

t

H

ow

hig

h a

bove t

he f

irst

flo

or

is t

he

seco

nd

flo

or?

3

. H

EX

AG

ON

S A

box o

f ch

oco

late

s sh

ap

ed

lik

e a

regu

lar

hexagon

is

pla

ced

sn

ugly

insi

de o

f a r

ect

an

gu

lar

box a

s sh

ow

n i

n

the f

igu

re.

If

th

e s

ide l

en

gth

of

the h

exagon

is

3 i

nch

es,

wh

at

are

th

e d

imen

sion

s of

the r

ect

an

gu

lar

box?

4

. W

IND

OW

S A

larg

e s

tain

ed

gla

ss

win

dow

is

con

stru

cted

fro

m s

ix 3

0°-

60°-

90°

tria

ngle

s as

show

n i

n t

he f

igu

re.

3 m

W

hat

is t

he h

eig

ht

of

the w

ind

ow

?

5

. M

OV

IES

K

im a

nd

Yola

nd

a a

re

watc

hin

g a

movie

in

a m

ovie

th

eate

r.

Yola

nd

a i

s si

ttin

g x

feet

from

th

e s

creen

an

d K

im i

s 15 f

eet

beh

ind

Yola

nd

a.

30

°45

°

15 f

tx

ft

T

he a

ngle

th

at

Kim

’s l

ine o

f si

gh

t

to t

he t

op

of

the s

creen

mak

es

wit

h

the h

ori

zon

tal

is 3

0°.

Th

e a

ngle

th

at

Yola

nd

a’s

lin

e o

f si

gh

t to

th

e t

op

of

the

scre

en

mak

es

wit

h t

he h

ori

zon

tal

is 4

5°.

a.

How

hig

h i

s th

e t

op

of

the s

creen

in

term

s of

x?

b.

Wh

at

is x

+ 1

5

−

x

?

c.

How

far

is Y

ola

nd

a f

rom

th

e s

creen

?

Rou

nd

you

r an

swer

to t

he n

eare

st

ten

th.

8-3

1

50 √

#

2 m

m

6

in

. b

y 3

√ #

3 i

n.

2

0ft

4 √

#

3

−3

m ≈

2.3

1 m

√ #

3 f

t

x f

t

20.5

ft

Lesson 8-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

23

G

len

co

e G

eo

me

try

En

ric

hm

en

t

√1

1

1

3√

2

Co

nstr

ucti

ng

Valu

es o

f S

qu

are

Ro

ots

Th

e d

iagra

m a

t th

e r

igh

t sh

ow

s a r

igh

t is

osc

ele

s tr

ian

gle

wit

h

two l

egs

of

len

gth

1 i

nch

. B

y t

he P

yth

agore

an

Th

eore

m,

the l

en

gth

of

the h

yp

ote

nu

se i

s √

$

2 i

nch

es.

By c

on

stru

ctin

g a

n a

dja

cen

t ri

gh

t

tria

ngle

wit

h l

egs

of

√ $

2

inch

es

an

d 1

in

ch,

you

can

cre

ate

a s

egm

en

t

of

len

gth

√ $

3 .

By c

on

tin

uin

g t

his

pro

cess

as

show

n b

elo

w,

you

can

con

stru

ct a

“wh

eel”

of

squ

are

roots

. T

his

wh

eel

is c

all

ed

th

e “

Wh

eel

of

Th

eod

oru

s”

aft

er

a G

reek

ph

iloso

ph

er

wh

o l

ived

abou

t 400 B

.C.

Con

tin

ue c

on

stru

ctin

g t

he w

heel

un

til

you

mak

e a

segm

en

t of

len

gth

√ $

$

18 .

1

1

11

1

1√ 2

√ 3

√5

√ 6

√ 7 √ 8

10

√$

√$

11

√$ 12

√$ 13

√$ 14

√$ 15

√$ 17

√$ 18

√$ 16

= 4

√ 4

= 2

√ 9

= 3

8-3


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

24

G

len

co

e G

eo

me

try

Tri

go

no

me

tric

Ra

tio

s T

he r

ati

o o

f th

e l

en

gth

s of

two s

ides

of

a r

igh

t tr

ian

gle

is

call

ed

a t

rig

on

om

etr

ic r

ati

o.

Th

e t

hre

e m

ost

com

mon

rati

os

are

sin

e,

co

sin

e,

an

d t

an

ge

nt,

wh

ich

are

abbre

via

ted

sin

, co

s, a

nd

ta

n,

resp

ect

ively

.

sin

R =

le

g o

pp

osi

te ∠

R

−

hyp

ote

nu

se

co

s R

=

leg a

dja

cen

t to

∠R

−

−

h

yp

ote

nu

se

ta

n R

=

leg o

pp

osi

te ∠

R

−

−

le

g a

dja

cen

t to

∠R

=

r −

t =

s −

t =

r −

s

F

ind

sin

A,

co

s A

, a

nd

ta

n A

. E

xp

re

ss e

ach

ra

tio

as

a f

ra

cti

on

an

d a

de

cim

al

to t

he

ne

are

st

hu

nd

re

dth

.

sin

A =

op

posi

te l

eg

−

hyp

ote

nu

se

cos

A =

ad

jace

nt

leg

−

h

yp

ote

nu

se

tan

A =

op

posi

te l

eg

−

ad

jace

nt

leg

=

BC

−

BA

=

AC

−

AB

=

BC

−

AC

=

5

−

13

= 1

2

−

13

=

5

−

12

≈ 0

.38

≈ 0

.92

≈ 0

.42

Exerc

ises

Fin

d s

in J

, co

s J

, ta

n J

, sin

L,

co

s L

, a

nd

ta

n L

. E

xp

re

ss e

ach

ra

tio

as a

fra

cti

on

an

d a

s a

de

cim

al

to t

he

ne

are

st

hu

nd

re

dth

if

ne

ce

ssa

ry

.

1.

2.

3.

1213

5 CB

A

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Trigonometry

8-4

s

tr TS

R

Exam

ple

20

12

16

40

24

32

36

12

√3

24

√3

s

in J

= 1

2

−

20 =

0.6

;

c

os

J =

16

−

20 =

0.8

;

ta

n J

= 1

2

−

16 =

0.7

5;

s

in L

= 1

6

−

20 =

0.8

;

c

os

L=

12

− 20

= 0

.6;

ta

n L

= 1

6

−

12 ≈

1.3

3

s

in J

= 2

4

−

40 =

0.6

;

c

os

J =

32

−

40 =

0.8

;

ta

n J

= 2

4

−

32 =

0.7

5;

s

in L

= 3

2

−

40 =

0.8

;

c

os

L =

24

−

32 =

0.6

;

ta

n L

= 3

2

−

24 ≈

1.3

3

s

in J

=

36

−

24 √

$

3

≈ 0

.87

;

c

os

J =

12 √

$

3

−

24 √

$

3

= 0

.5;

ta

n J

=

36

−

12 √

$

3

≈ 1

.73

;

s

in L

= 1

2 √

$

3

−

24 √

$

3

= 0

.5;

c

os

L =

36

−

24 √

$

3

≈ 0

.87

;

ta

n L

= 1

2 √

$

3

−

36

≈ 0

.58


NA

ME

DA

TE

PE

RIO

D

Lesson 8-4

Ch

ap

ter

8

25

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Trigonometry

Use

In

ve

rse

Tri

go

no

me

tric

Ra

tio

s Y

ou

can

use

a c

alc

ula

tor

an

d t

he s

ine,

cosi

ne,

or

tan

gen

t to

fin

d t

he m

easu

re o

f th

e a

ngle

, ca

lled

th

e i

nv

erse

of

the t

rigon

om

etr

ic r

ati

o.

U

se

a c

alc

ula

tor t

o f

ind

th

e m

ea

su

re

of

∠T

to

th

e n

ea

re

st

ten

th.

Th

e m

easu

res

giv

en

are

th

ose

of

the l

eg o

pp

osi

te ∠

T a

nd

th

e

hyp

ote

nu

se,

so w

rite

an

equ

ati

on

usi

ng t

he s

ine r

ati

o.

sin

T =

op

p

−

hyp

sin

T =

2

9

−

34

If s

in T

= 2

9

−

34 ,

then

sin

-1 2

9

−

34 =

m∠

T.

Use

a c

alc

ula

tor.

So,

m∠

T ≈

58.5

.

Exerc

ises

Use

a c

alc

ula

tor t

o f

ind

th

e m

ea

su

re

of

∠T

to

th

e n

ea

re

st

ten

th.

1.

34

14

√3

2.

7

18

3.

87

34

4.

101

32

5.

67

10

√3

6.

39

14

√2

8-4 Exam

ple

34

29

71.5

°75°

30.5

°

35.5

°22.9

°67°


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

26

G

len

co

e G

eo

me

try

Fin

d s

in R

, co

s R

, ta

n R

, sin

S,

co

s S

, a

nd

ta

n S

.

Ex

pre

ss e

ach

ra

tio

as a

fra

cti

on

an

d a

s a

de

cim

al

to t

he

ne

are

st

hu

nd

re

dth

.

1. r =

16, s =

30, t

= 3

4

2. r =

10, s =

24, t

= 2

6

Use

a s

pe

cia

l rig

ht

tria

ng

le t

o e

xp

re

ss e

ach

trig

on

om

etr

ic r

ati

o a

s a

fra

cti

on

an

d

as a

de

cim

al

to t

he

ne

are

st

hu

nd

re

dth

if

ne

ce

ssa

ry

.

3. si

n 3

0°

4. ta

n 4

5°

5. co

s 60°

6. si

n 6

0°

7. ta

n 3

0°

8. co

s 45°

Fin

d x

. R

ou

nd

to

th

e n

ea

re

st

hu

nd

re

dth

if

ne

ce

ssa

ry

.

9.

7x

36°

10.

11.

Use

a c

alc

ula

tor t

o f

ind

th

e m

ea

su

re

of ∠B

to

th

e n

ea

re

st

ten

th.

12.

13.

14.

sR

S Trt

Sk

ills

Practi

ce

Trigonometry

8-4

12

x

15°

6

x°

18

15

x°

12

22

x°

19

12

x

63°

sin

R =

8

−

17 ≈

0.4

7;

sin

R =

5

−

13 ≈

0.3

8;

co

s R

= 1

5

−

17 ≈

0.8

8;

co

s R

= 1

2

−

13 ≈

0.9

2;

tan

R =

8

−

15 ≈

0.5

3;

ta

n R

=

5

−

12 ≈

0.4

2;

sin

S =

15

−

17 ≈

0.8

8;

sin

S =

12

−

13 ≈

0.9

2;

co

s S

=

8

−

17 ≈

0.4

7;

co

s S

=

5

−

13 ≈

0.3

8;

tan

S =

15

−

8

≈ 1

.88

tan

S =

12

−

5

≈ 2

.4

x =

5.0

9x =

13.4

7x =

3.1

1

49.2

°70.5

°36.9

°

1

−

2 ;

0.5

1 1

−

2 ;

0.5

√ $

3

−

2

; 0.8

7

√ $

3

−

3

; 0.5

8

√ $

2

−

2

; 0.7

1

Lesson 8-4


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

27

G

len

co

e G

eo

me

try

Fin

d s

in L

, co

s L

, ta

n L

, sin

M,

co

s M

, a

nd

ta

n M

.

Ex

pre

ss e

ach

ra

tio

as a

fra

cti

on

an

d a

s a

de

cim

al

to t

he

ne

are

st

hu

nd

re

dth

.

1.

ℓ =

15, m

= 3

6, n

= 3

9

2.

ℓ =

12, m

= 1

2 √

#

3 , n

= 2

4

Fin

d x

. R

ou

nd

to

th

e n

ea

re

st

hu

nd

re

dth

.

3

.

11

64

°

x

4

.

5.

41

°x

32

Use

a c

alc

ula

tor t

o f

ind

th

e m

ea

su

re

of

∠B

to

th

e n

ea

re

st

ten

th.

6

.

8

14

7

.

30

25

8

.

9

. G

EO

GR

AP

HY

D

iego u

sed

a t

heod

oli

te t

o m

ap

a r

egio

n o

f la

nd

for

his

class

in

geom

orp

holo

gy.

To d

ete

rmin

e t

he e

levati

on

of

a v

ert

ical

rock

form

ati

on

, h

e m

easu

red

th

e d

ista

nce

fro

m t

he b

ase

of

the f

orm

ati

on

to

his

posi

tion

an

d t

he a

ngle

betw

een

th

e g

rou

nd

an

d t

he l

ine o

f si

gh

t to

th

e

top

of

the f

orm

ati

on

. T

he d

ista

nce

was

43 m

ete

rs a

nd

th

e a

ngle

was

36°.

Wh

at

is t

he h

eig

ht

of

the f

orm

ati

on

to t

he n

eare

st m

ete

r?

29

29

°x

36

° 43m

ML

N

m

n

Practi

ce

Trigonometry

8-4

739

sin

L =

5

−

13 ≈

0.3

8;

sin

L =

1

−

2 =

0.5

0;

co

s L

= 1

2

−

13 ≈

0.9

2;

co

s L

= √

$

3

−

2 ≈

0.8

7;

tan

L =

5

−

12 ≈

0.4

2;

ta

n L

=

1

−

√ $

3

or

√ $

3

−

3 ≈

0.5

8;

sin

M =

12

−

13 ≈

0.9

2;

sin

M =

√ $

3

−

2 ≈

0.8

7;

co

s M

=

5

−

13 ≈

0.3

8;

co

s M

= 1

−

2 =

0.5

0;

tan

M =

12

−

5

≈ 2

.4

tan

M =

√ $

3 ≈

1.7

3

60.3

°33.6

°79.7

°

22.5

525.3

624.1

5

31 m


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

28

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

Tri

go

no

metr

y

1

. R

AD

IO T

OW

ER

S K

ay i

s st

an

din

g n

ear

a 2

00-f

oot-

hig

h r

ad

io t

ow

er.

200 f

t ?

ft

49

°

U

se t

he i

nfo

rmati

on

in

th

e f

igu

re t

o

dete

rmin

e h

ow

far

Kay i

s fr

om

th

e t

op

of

the t

ow

er.

Exp

ress

you

r an

swer

as

a

trig

on

om

etr

ic f

un

ctio

n.

2

. R

AM

PS

A

60-f

oot

ram

p r

ises

from

th

e

firs

t fl

oor

to t

he s

eco

nd

flo

or

of

a p

ark

ing

gara

ge.

Th

e r

am

p m

ak

es

a 1

5°

an

gle

w

ith

th

e g

rou

nd

.

? f

t

60 f

t

15

°

H

ow

hig

h a

bove t

he f

irst

flo

or

is t

he

seco

nd

flo

or?

Exp

ress

you

r an

swer

as

a

trig

on

om

etr

ic f

un

ctio

n.

3

. T

RIG

ON

OM

ET

RY

M

eli

nd

a a

nd

Walt

er

were

both

solv

ing t

he s

am

e t

rigon

om

etr

y

pro

ble

m.

How

ever,

aft

er

they f

inis

hed

th

eir

com

pu

tati

on

s, M

eli

nd

a s

aid

th

e

an

swer

was

52 s

in 2

7°

an

d W

alt

er

said

th

e a

nsw

er

was

52 c

os

63

°. C

ou

ld t

hey

both

be c

orr

ect

? E

xp

lain

.

4

. LIN

ES

Jasm

ine d

raw

s li

ne m

on

a

coord

inate

pla

ne.

m x

y

O-5

5

5

W

hat

an

gle

does

m m

ak

e w

ith

th

e

x-a

xis

? R

ou

nd

you

r an

swer

to t

he

neare

st d

egre

e.

5

. N

EIG

HB

OR

S A

my,

Barr

y,

an

d C

hri

s li

ve

on

th

e s

am

e b

lock

. C

hri

s li

ves

up

th

e

stre

et

an

d a

rou

nd

th

e c

orn

er

from

Am

y,

an

d B

arr

y l

ives

at

the c

orn

er

betw

een

A

my a

nd

Ch

ris.

Th

e t

hre

e h

om

es

are

th

e

vert

ices

of

a r

igh

t tr

ian

gle

.

64

°

Bar

ryChri

s

Am

y

26

°

a.

Giv

e t

wo t

rigon

om

etr

ic e

xp

ress

ion

s fo

r th

e r

ati

o o

f B

arr

y’s

dis

tan

ce f

rom

A

my t

o C

hri

s’ d

ista

nce

fro

m A

my.

b.

Giv

e t

wo t

rigon

om

etr

ic e

xp

ress

ion

s fo

r th

e r

ati

o o

f B

arr

y’s

dis

tan

ce f

rom

C

hri

s to

Am

y’s

dis

tan

ce f

rom

Ch

ris.

c.

Giv

e a

tri

gon

om

etr

ic e

xp

ress

ion

for

the r

ati

o o

f A

my’s

dis

tan

ce f

rom

B

arr

y t

o C

hri

s’ d

ista

nce

fro

m B

arr

y.

8-4

20

0

−

sin

49

°

60 s

in 1

5°

Yes,

they a

re b

oth

co

rrect.

B

ecau

se 2

7 +

63

= 9

0,

the

sin

e o

f 27°

is t

he s

am

e r

ati

o a

s

the c

osin

e o

f 63°.

co

s 2

6°

or

sin

64°

co

s 6

4°

or

sin

26°

tan

64°

18°


NA

ME

DA

TE

PE

RIO

D

Lesson 8-4

Ch

ap

ter

8

29

G

len

co

e G

eo

me

try

En

ric

hm

en

t

Sin

e a

nd

Co

sin

e o

f A

ng

les

Th

e f

oll

ow

ing d

iagra

m c

an

be u

sed

to o

bta

in a

pp

roxim

ate

valu

es

for

the s

ine a

nd

cosi

ne o

f an

gle

s fr

om

0°

to 9

0°.

Th

e r

ad

ius

of

the c

ircl

e i

s 1.

So,

the s

ine a

nd

cosi

ne v

alu

es

can

be

read

dir

ect

ly f

rom

th

e v

ert

ical

an

d h

ori

zon

tal

axes.

F

ind

ap

pro

xim

ate

va

lue

s f

or s

in 4

0°

an

d

co

s 4

0°.

Co

nsid

er t

he

tria

ng

le f

orm

ed

by

th

e s

eg

me

nt

ma

rk

ed

40

°, a

s i

llu

str

ate

d b

y t

he

sh

ad

ed

tria

ng

le a

t rig

ht.

sin

40°

= a

−

c ≈

0.6

4

−

1

or

0.6

4

cos

40°

= b

−

c ≈

0.7

7

−

1

or

0.7

7

1

. U

se t

he d

iagra

m a

bove t

o c

om

ple

te t

he c

hart

of

valu

es.

x°

0°

10

°20

°30°

40

°50

°60

°70

°80

°90

°

sin x

°0

0.1

70.3

40.5

0.6

40.7

70.8

70.9

40.9

81

cos x

°1

0.9

80.9

40.8

70.7

70.6

40.5

0.3

40.1

70

2

. C

om

pare

th

e s

ine a

nd

cosi

ne o

f tw

o c

om

ple

men

tary

an

gle

s (a

ngle

s w

ith

a

sum

of

90°)

. W

hat

do y

ou

noti

ce?

00.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.91

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

90

°

0°

10

°

20

°

30

°

40

°

50

°

60

°

70

°

80

°

1 0

40

°0

.64

c=

1 u

nit

x°

b=

co

s x°

0.7

71

a=

sin

x°

Exam

ple

8-4 T

he s

ine o

f an

an

gle

is e

qu

al

to t

he c

osin

e o

f th

e

co

mp

lem

en

t o

f th

e a

ng

le.


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

30

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

An

gle

s o

f E

levati

on

an

d D

ep

ressio

n

An

gle

s o

f E

lev

ati

on

an

d D

ep

ress

ion

M

an

y r

eal-

worl

d p

roble

ms

that

involv

e l

ook

ing u

p t

o a

n o

bje

ct c

an

be d

esc

ribed

in

term

s of

an

an

gle

o

f e

lev

ati

on

, w

hic

h i

s th

e a

ngle

betw

een

an

obse

rver’

s li

ne o

f si

gh

t an

d a

hori

zon

tal

lin

e.

Wh

en

an

obse

rver

is l

ook

ing d

ow

n,

the a

ng

le o

f d

ep

re

ssio

n i

s th

e

an

gle

betw

een

th

e o

bse

rver’

s li

ne o

f si

gh

t an

d a

hori

zon

tal

lin

e.

T

he

an

gle

of

ele

va

tio

n f

ro

m p

oin

t A

to

th

e t

op

of

a c

liff

is 3

4°.

If

po

int A

is 1

00

0 f

ee

t fr

om

th

e b

ase

of

the

cli

ff,

ho

w h

igh

is t

he

cli

ff?

Let

x =

th

e h

eig

ht

of

the c

liff

.

ta

n 3

4°

=

x

−

1000

tan =

opposite

−

adja

cent

1000(t

an

34°)

= x

M

ultip

ly e

ach s

ide b

y 1

000.

674.5

≈ x

U

se a

calc

ula

tor.

Th

e h

eig

ht

of

the c

liff

is

abou

t 674.5

feet.

Exerc

ises

1

. H

ILL T

OP

T

he a

ngle

of

ele

vati

on

fro

m p

oin

t A

to t

he t

op

of

a h

ill

is 4

9°.

If

poin

t A

is

400 f

eet

from

th

e b

ase

of

the h

ill,

how

hig

h i

s th

e h

ill?

2. S

UN

F

ind

th

e a

ngle

of

ele

vati

on

of

the S

un

wh

en

a 1

2.5

-mete

r-ta

ll

tele

ph

on

e p

ole

cast

s an

18-m

ete

r-lo

ng s

had

ow

.

3. S

KII

NG

A

sk

i ru

n i

s 1000 y

ard

s lo

ng w

ith

a v

ert

ical

dro

p o

f 208 y

ard

s. F

ind

th

e a

ngle

of

dep

ress

ion

fro

m t

he t

op

of

the s

ki

run

to t

he b

ott

om

.

4

. A

IR T

RA

FFIC

F

rom

th

e t

op

of

a 1

20-f

oot-

hig

h t

ow

er,

an

air

tra

ffic

con

troll

er

obse

rves

an

air

pla

ne o

n t

he r

un

way

at

an

an

gle

of

dep

ress

ion

of

19°.

How

far

from

th

e b

ase

of

the t

ow

er

is t

he a

irp

lan

e?

18 m

12.5

mSun

?

400 f

t

?

49°

A

?

1000 f

t

34°

x

angle

of

elev

atio

n

line

of s

ight

8-5 Exam

ple

Ylin

e of

sig

ht

hori

zonta

lan

gle

of

dep

ress

ion

120 f

t

?

19°

208 y

d

?

1000 y

d

460 f

t

35°

12°

348.5

ft

Lesson 8-5


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

31

G

len

co

e G

eo

me

try

Tw

o A

ng

les

of

Ele

va

tio

n o

r D

ep

ress

ion

A

ngle

s of

ele

vati

on

or

dep

ress

ion

to t

wo

dif

fere

nt

obje

cts

can

be u

sed

to e

stim

ate

dis

tan

ce b

etw

een

th

ose

obje

cts.

Th

e a

ngle

s fr

om

tw

o d

iffe

ren

t p

osi

tion

s of

obse

rvati

on

to t

he s

am

e o

bje

ct c

an

be u

sed

to e

stim

ate

th

e h

eig

ht

of

the o

bje

ct.

T

o e

sti

ma

te t

he

he

igh

t o

f a

ga

ra

ge

,

Ja

so

n s

igh

ts t

he

to

p o

f th

e g

ara

ge

at

a 4

2°

an

gle

of

ele

va

tio

n.

He

th

en

ste

ps b

ack

20

fe

et

an

d s

ite

s

the

to

p a

t a

10

° a

ng

le.

If J

aso

n i

s 6

fe

et

tall

,

ho

w t

all

is t

he

ga

ra

ge

to

th

e n

ea

re

st

foo

t?

△ A

BC

an

d △

AB

D a

re r

igh

t tr

ian

gle

s. W

e c

an

dete

rmin

e A

B =

x a

nd

CB

= y

, an

d D

B =

y +

20.

Use

△ A

BC

. U

se △

AB

D.

tan

42°

= x

−

y o

r y t

an

42°

= x

ta

n 1

0°

=

x

−

y +

20 o

r (y

+ 2

0)

tan

10°

= x

Su

bst

itu

te t

he v

alu

e f

or

x f

rom

△ A

BD

in

th

e e

qu

ati

on

for

△ A

BC

an

d s

olv

e f

or

y.

y t

an

42°

= (

y +

20)

tan

10°

y t

an

42°

= y

tan

10°

+ 2

0 t

an

10°

y t

an

42°

− y

tan

10°

= 2

0 t

an

10°

y (

tan

42°

− t

an

10°)

= 2

0 t

an

10°

y =

20 t

an

10°

−

−

ta

n 4

2°

- t

an

10° ≈

4.8

7

If y

≈ 4

.87,

then

x =

4.8

7 t

an

42°

or

abou

t 4.4

feet.

Ad

d J

aso

n’s

heig

ht,

so t

he g

ara

ge i

s abou

t 4.4

+ 6

or

10.4

feet

tall

.

Exerc

ises

1

. C

LIF

F S

ara

h s

tan

ds

on

th

e g

rou

nd

an

d s

igh

ts t

he t

op

of

a s

teep

cli

ff a

t a 6

0°

an

gle

of

ele

vati

on

. S

he t

hen

st

ep

s back

50 m

ete

rs a

nd

sig

hts

th

e t

op

of

the s

teep

cl

iff

at

a 3

0°

an

gle

. If

Sara

h i

s 1.8

mete

rs t

all

, h

ow

tall

is

the s

teep

cli

ff t

o t

he n

eare

st m

ete

r?

2. B

ALLO

ON

T

he a

ngle

of

dep

ress

ion

fro

m a

hot

air

ball

oon

in

th

e a

ir t

o a

pers

on

on

th

e g

rou

nd

is

36°.

If

the p

ers

on

st

ep

s back

10 f

eet,

th

e n

ew

an

gle

of

dep

ress

ion

is

25°.

If

th

e p

ers

on

is

6 f

eet

tall

, h

ow

far

off

th

e g

rou

nd

is

the

hot

air

ball

oon

?

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

An

gle

s o

f E

levati

on

an

d D

ep

ressio

n

8-5 Exam

ple

42°

10°

x

y6

ft

20 f

t

Bu

ildin

g

60°

30°

x

y1

.8m

50m

Ste

epcl

iff

36°

25°

x

y6

ft

10

ft

Ballo

on

ab

ou

t 45 m

ete

rs

ab

ou

t 19 f

eet


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

32

G

len

co

e G

eo

me

try

Na

me

th

e a

ng

le o

f d

ep

re

ssio

n o

r a

ng

le o

f e

lev

ati

on

in

ea

ch

fig

ure

.

1

.

2.

3

.

4.

5. M

OU

NT

AIN

BIK

ING

O

n a

mou

nta

in b

ike t

rip

alo

ng t

he G

em

ini

Bri

dges

Tra

il i

n M

oab,

Uta

h,

Nabu

ko s

top

ped

on

th

e c

an

yon

flo

or

to g

et

a g

ood

vie

w o

f th

e t

win

san

dst

on

e

bri

dges.

Nabu

ko i

s st

an

din

g a

bou

t 60 m

ete

rs f

rom

th

e b

ase

of

the c

an

yon

cli

ff,

an

d t

he

natu

ral

arc

h b

rid

ges

are

abou

t 100 m

ete

rs u

p t

he c

an

yon

wall

. If

her

lin

e o

f si

gh

t is

5 m

etr

es

above t

he g

rou

nd

, w

hat

is t

he a

ngle

of

ele

vati

on

to t

he t

op

of

the b

rid

ges?

Rou

nd

to t

he n

eare

st t

en

th d

egre

e.

6. S

HA

DO

WS

S

up

pose

th

e s

un

cast

s a s

had

ow

off

a 3

5-f

oot

bu

ild

ing.

If t

he a

ngle

of

ele

vati

on

to t

he s

un

is

60°,

how

lon

g i

s th

e s

had

ow

to t

he n

eare

st t

en

th o

f a f

oot?

7

. B

ALLO

ON

ING

A

ngie

sees

a h

ot

air

ball

oon

in

th

e

sky f

rom

her

spot

on

th

e g

rou

nd

. T

he a

ngle

of

ele

vati

on

fro

m A

ngie

to t

he b

all

oon

is

40°.

If

she s

tep

s

back

200 f

eet,

th

e n

ew

an

gle

of

ele

vati

on

is

10°.

If A

ngie

is

5.5

feet

tall

, h

ow

far

off

th

e g

rou

nd

is t

he h

ot

air

ball

oon

?

8. IN

DIR

EC

T M

EA

SU

RE

ME

NT

K

yle

is

at

the e

nd

of

a p

ier

30 f

eet

above t

he o

cean

. H

is e

ye l

evel

is

3 f

eet

above t

he p

ier.

He i

s u

sin

g b

inocu

lars

to

watc

h a

wh

ale

su

rface

. If

th

e a

ngle

of

dep

ress

ion

of

the w

hale

is

20°,

how

far

is t

he w

hale

fro

m

Kyle

’s b

inocu

lars

? R

ou

nd

to t

he n

eare

st t

en

th f

oot.

whal

ew

ater

leve

l

20°

Kyl

e’s

eyes

pie

r3 f

t

30 f

t

60° ?

35 f

t

Z

PW

R

D

AC

B

T

WR

S

F

T

L

S

Sk

ills

Practi

ce

An

gle

s o

f E

levati

on

an

d D

ep

ressio

n

8-5

40°

10°

x

y5.5

ft

200 f

t

Bal

loon

ab

ou

t 50.1

ft

ab

ou

t 96.5

ft

ab

ou

t 20.2

ft

ab

ou

t 57.7

°

∠FLS

; ∠

TS

L∠

RTW

; ∠

SW

T

∠D

CB

; ∠

AB

C∠

WZ

P; ∠

RP

Z

Lesson 8-5


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

33

G

len

co

e G

eo

me

try

Na

me

th

e a

ng

le o

f d

ep

re

ssio

n o

r a

ng

le o

f e

lev

ati

on

in

ea

ch

fig

ure

.

1.

2.

3. W

AT

ER

TO

WE

RS

A

stu

den

t ca

n s

ee a

wate

r to

wer

from

th

e c

lose

st p

oin

t of

the s

occ

er

field

at

San

Lobos

Hig

h S

chool.

Th

e e

dge o

f th

e s

occ

er

field

is

abou

t 110 f

eet

from

th

e

wate

r to

wer

an

d t

he w

ate

r to

wer

stan

ds

at

a h

eig

ht

of

32.5

feet.

Wh

at

is t

he a

ngle

of

ele

vati

on

if

the e

ye l

evel

of

the s

tud

en

t vie

win

g t

he t

ow

er

from

th

e e

dge o

f th

e s

occ

er

field

is

6 f

eet

above t

he g

rou

nd

? R

ou

nd

to t

he n

eare

st t

en

th.

4. C

ON

ST

RU

CT

ION

A

roofe

r p

rop

s a l

ad

der

again

st a

wall

so t

hat

the t

op

of

the l

ad

der

reach

es

a 3

0-f

oot

roof

that

need

s re

pair

. If

th

e a

ngle

of

ele

vati

on

fro

m t

he b

ott

om

of

the

lad

der

to t

he r

oof

is 5

5°,

how

far

is t

he l

ad

der

from

th

e b

ase

of

the w

all

? R

ou

nd

you

r

an

swer

to t

he n

eare

st f

oot.

5. T

OW

N O

RD

INA

NC

ES

T

he t

ow

n o

f B

elm

on

t re

stri

cts

the h

eig

ht

of

flagp

ole

s to

25 f

eet

on

an

y p

rop

ert

y.

Lin

dsa

y w

an

ts t

o

dete

rmin

e w

heth

er

her

sch

ool

is i

n c

om

pli

an

ce w

ith

th

e r

egu

lati

on

.

Her

eye l

evel

is 5

.5 f

eet

from

th

e g

rou

nd

an

d s

he s

tan

ds

36 f

eet

from

th

e f

lagp

ole

. If

th

e a

ngle

of

ele

vati

on

is

abou

t 25°,

wh

at

is t

he h

eig

ht

of

the f

lagp

ole

to t

he n

eare

st t

en

th?

6. G

EO

GR

APH

Y S

tep

han

is

stan

din

g o

n t

he g

rou

nd

by

a m

esa

in

th

e P

ain

ted

Dese

rt.

Ste

ph

an

is

1.8

mete

rs

tall

an

d s

igh

ts t

he t

op

of

the m

esa

at

29°.

Ste

ph

an

ste

ps

back

100 m

ete

rs a

nd

sig

hts

th

e t

op

at

25°.

How

tall

is

the m

esa

?

7. IN

DIR

EC

T M

EA

SU

RE

ME

NT

M

r. D

om

ingu

ez i

s st

an

din

g

on

a 4

0-f

oot

oce

an

blu

ff n

ear

his

hom

e.

He c

an

see h

is t

wo

dogs

on

th

e b

each

belo

w.

If h

is l

ine o

f si

gh

t is

6 f

eet

above

the g

rou

nd

an

d t

he a

ngle

s of

dep

ress

ion

to h

is d

ogs

are

34°

an

d 4

8°,

how

far

ap

art

are

th

e d

ogs

to t

he n

eare

st f

oot?

48°

34°

40

ft

6 f

t

Mr.

Do

min

gu

ez

blu

ff

25°

5.5

ft

36

ft

x

R

M

P

L

T

YR

ZY

Practi

ce

An

gle

s o

f E

levati

on

an

d D

ep

ressio

n

8-5

29°

25°

x

y1

.8 m

10

0 m

mes

a

ab

ou

t 21 f

t

ab

ou

t 22.3

ft

ab

ou

t 27 f

t

ab

ou

t 296 m

ab

ou

t 13.5

°

∠TR

Z; ∠

YZ

R∠

PR

M; ∠

LM

R


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

34

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

An

gle

s o

f E

levati

on

an

d D

ep

ressio

n

1

. LIG

HT

HO

US

ES

S

ail

ors

on

a s

hip

at

sea

spot

the l

igh

t fr

om

a l

igh

thou

se.

Th

e

an

gle

of

ele

vati

on

to t

he l

igh

t is

25°.

25°

30m

?

T

he l

igh

t of

the l

igh

thou

se i

s 30 m

ete

rs

above s

ea l

evel.

How

far

from

th

e s

hore

is t

he s

hip

? R

ou

nd

you

r an

swer

to t

he

neare

st m

ete

r.

2

. R

ES

CU

E A

hik

er

dro

pp

ed

his

back

pack

over

on

e s

ide o

f a c

an

yon

on

to a

led

ge

belo

w.

Beca

use

of

the s

hap

e o

f th

e c

liff

,

he c

ou

ld n

ot

see e

xact

ly w

here

it

lan

ded

.

32°

115 f

t

? f

t

Bac

kpac

k

F

rom

th

e o

ther

sid

e,

the p

ark

ran

ger

rep

ort

s th

at

the a

ngle

of

dep

ress

ion

to

the b

ack

pack

is

32°.

If

the w

idth

of

the

can

yon

is

115 f

eet,

how

far

dow

n d

id

the b

ack

pack

fall

? R

ou

nd

you

r an

swer

to t

he n

eare

st f

oot.

3

. A

IRP

LA

NE

S T

he a

ngle

of

ele

vati

on

to

an

air

pla

ne v

iew

ed

fro

m t

he c

on

trol

tow

er

at

an

air

port

is

7°.

Th

e t

ow

er

is

200 f

eet

hig

h a

nd

th

e p

ilot

rep

ort

s th

at

the a

ltit

ud

e i

s 5200 f

eet.

How

far

aw

ay

from

th

e c

on

trol

tow

er

is t

he a

irp

lan

e?

Rou

nd

you

r an

swer

to t

he n

eare

st f

oot.

4

. P

EA

K T

RA

M T

he P

eak

Tra

m i

n H

on

g

Kon

g c

on

nect

s tw

o t

erm

inals

, on

e a

t th

e

base

of

a m

ou

nta

in,

an

d t

he o

ther

at

the

sum

mit

. T

he a

ngle

of

ele

vati

on

of

the

up

per

term

inal

from

th

e l

ow

er

term

inal

is a

bou

t 15.5

°. T

he d

ista

nce

betw

een

th

e

two t

erm

inals

is

abou

t 1365 m

ete

rs.

Abou

t h

ow

mu

ch h

igh

er

above s

ea l

evel

is t

he u

pp

er

term

inal

com

pare

d t

o t

he

low

er

term

inal?

Rou

nd

you

r an

swer

to

the n

eare

st m

ete

r.

5

. H

ELIC

OP

TE

RS

Jerm

ain

e a

nd

Joh

n a

re

watc

hin

g a

heli

cop

ter

hover

above t

he

gro

un

d.

48°

55°

(Not

dra

wn t

o s

cale

)

Hel

icopte

r

Jerm

aine

John

10 m

h

x

Jerm

ain

e a

nd

Joh

n a

re s

tan

din

g

10 m

ete

rs a

part

.

a.

Fin

d t

wo d

iffe

ren

t exp

ress

ion

s th

at

can

be u

sed

to f

ind

th

e h

, h

eig

ht

of

the h

eli

cop

ter.

b.

Equ

ate

th

e t

wo e

xp

ress

ion

s you

fou

nd

for

Exerc

ise a

to s

olv

e f

or x.

Rou

nd

you

r an

swer

to t

he n

eare

st

hu

nd

red

th.

c.

How

hig

h a

bove t

he g

rou

nd

is

the

heli

cop

ter?

Rou

nd

you

r an

swer

to t

he

neare

st h

un

dre

dth

.

8-5 6

4 m

72 f

t

41,0

28 f

t

365 m

h =

x t

an

55°;

h

= (

x +

10)

tan

48°

34.9

8 m

49.9

5 m

Lesson 8-5


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

35

G

len

co

e G

eo

me

try

En

ric

hm

en

t

Best

Seat

in t

he H

ou

se

Most

peop

le w

an

t to

sit

in

th

e b

est

seat

in t

he m

ovie

th

eate

r.

Th

e b

est

seat

cou

ld b

e

defi

ned

as

the s

eat

that

all

ow

s you

to s

ee t

he m

axim

um

am

ou

nt

of

scre

en

. T

he p

ictu

re

belo

w r

ep

rese

nts

th

is s

itu

ati

on

.

To d

ete

rmin

e t

he b

est

seat

in t

he h

ou

se,

you

wan

t to

fin

d w

hat

valu

e o

f x a

llow

s you

to s

ee

the m

axim

um

am

ou

nt

of

scre

en

. T

he v

alu

e o

f x i

s h

ow

far

from

th

e s

creen

you

sh

ou

ld s

it.

1

. T

o m

axim

ize t

he a

mou

nt

of

scre

en

vie

wed

, w

hic

h a

ngle

valu

e n

eed

s to

be m

axim

ized

?

Wh

y?

2

. W

hat

is t

he v

alu

e o

f a

if x =

10 f

eet?

3

. W

hat

is t

he v

alu

e o

f a

if x =

20 f

eet?

4

. W

hat

is t

he v

alu

e o

f a

if x =

25 f

eet?

5

. W

hat

is t

he v

alu

e o

f a

if x =

35 f

eet?

6

. W

hat

is t

he v

alu

e o

f a

if x =

55 f

eet?

7. W

hic

h v

alu

e o

f x g

ives

the g

reate

st v

alu

e o

f a

? S

o,

wh

ere

is

the b

est

seat

in t

he m

ovie

theate

r?

a ˚

b˚

c ˚ xft

scre

en

40

ft

10

ft

8-5 A

ng

le w

ith

measu

re a

becau

se t

his

is h

ow

mu

ch

of

the s

cre

en

can

b

e v

iew

ed

.

33.7

ft

41.6

ft

41.6

ft

39.1

ft

32 f

t

Th

e b

est

seat

in t

he h

ou

se i

s a

rou

nd

20–25 f

eet

aw

ay f

rom

th

e s

cre

en

.


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

36

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Th

e L

aw

of

Sin

es a

nd

Law

of

Co

sin

es

Th

e L

aw

of S

ine

s

In a

ny t

rian

gle

, th

ere

is

a s

peci

al

rela

tion

ship

betw

een

th

e a

ngle

s of

the t

rian

gle

an

d t

he l

en

gth

s of

the s

ides

op

posi

te t

he a

ngle

s.

F

ind

b.

Ro

un

d t

o t

he

ne

are

st

ten

th.

45

°

30

74

°

b

B

AC

sin

C

−

c

= si

n B

−

b

Law

of

Sin

es

si

n 4

5°

−

30

=

sin

74°

−

b

m

∠C

= 4

5,

c =

30,

m∠

B =

74

b s

in 4

5°

= 3

0 s

in 7

4°

Cro

ss P

roducts

Pro

pert

y

b =

30 s

in 7

4°

−

sin

45

°

Div

ide e

ach s

ide b

y s

in 4

5°.

b ≈

40.8

U

se a

calc

ula

tor.

F

ind

d.

Ro

un

d t

o t

he

ne

are

st

ten

th.

By t

he T

rian

gle

An

gle

-Su

m T

heore

m,

m∠

E =

180 -

(82 +

40)

or

58.

82

°40

°

24

d

si

n D

−

d

= si

n E

−

e

Law

of

Sin

es

si

n 8

2°

−

d

= si

n 5

8°

−

24

m

∠D

= 8

2,

m∠

E =

58,

e =

24

24 s

in 8

2°

= d

sin

58°

Cro

ss P

roducts

Pro

pert

y

24 s

in 8

2°

−

sin

58°

= d

D

ivid

e e

ach s

ide b

y s

in 5

8°.

d

≈ 2

8.0

U

se a

calc

ula

tor.

Exerc

ises

Fin

d x

. R

ou

nd

to

th

e n

ea

re

st

ten

th.

1.

80°

40°

x

12

2

.

52°

91°

x

20

3

.

16

x

84°

34°

4

.

17°

72°

x25

5

.

89°

80°

x12

6.

60°

35°

35

x

8-6

Exam

ple

1Exam

ple

2

La

w o

f S

ine

s s

in A

−

a

= s

in B

−

b

= s

in C

−

c

7.7

12.2

30.4

18.4

26.2

18.0

Lesson 8-6


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

37

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Th

e L

aw

of

Sin

es a

nd

Law

of

Co

sin

es

Th

e L

aw

of C

osin

es

An

oth

er

rela

tion

ship

betw

een

th

e s

ides

an

d a

ngle

s of

an

y t

rian

gle

is

call

ed

th

e L

aw

of

Co

sin

es.

You

can

use

th

e L

aw

of

Cosi

nes

if y

ou

kn

ow

th

ree s

ides

of

a

tria

ngle

or

if y

ou

kn

ow

tw

o s

ides

an

d t

he i

ncl

ud

ed

an

gle

of

a t

rian

gle

.

La

w o

f C

os

ine

s

Le

t △

AB

C b

e a

ny t

ria

ng

le w

ith

a,

b,

an

d c

re

pre

se

ntin

g t

he

me

asu

res o

f th

e s

ide

s o

pp

osite

the

an

gle

s w

ith

me

asu

res A

, B

, a

nd

C,

resp

ective

ly.

Th

en

th

e f

ollo

win

g e

qu

atio

ns a

re t

rue

.

a

2 =

b2 +

c2 -

2b

c c

os A

b

2 =

a2 -

c2 -

2a

c c

os B

c

2 =

a2 +

b2 -

2a

b c

os C

F

ind

c.

Ro

un

d t

o t

he

ne

are

st

ten

th.

c

2 =

a2 +

b2 -

2a

b c

os

C

Law

of

Cosin

es

c

2 =

12

2 +

10

2 -

2(1

2)(

10)c

os

48

° a

= 1

2,

b =

10,

m∠

C =

48

c =

√ '

''

''

''

''

''

'

12

2 +

10

2 -

2(1

2)(

10)c

os

48

° T

ake t

he s

quare

root

of

each s

ide.

c ≈

9.1

U

se a

calc

ula

tor.

F

ind

m∠A

. R

ou

nd

to

th

e n

ea

re

st

de

gre

e.

a

2 =

b2 +

c2 -

2bc c

os

A

Law

of

Cosin

es

7

2 =

52 +

82 -

2(5

)(8)

cos

A

a =

7,

b =

5,

c =

8

49 =

25 +

64 -

80 c

os

A

Multip

ly.

-

40 =

-80 c

os

A

Subtr

act

89 f

rom

each s

ide.

1

−

2 =

cos

A

Div

ide e

ach s

ide b

y -

80.

cos-

1 1

−

2 =

A

Use t

he invers

e c

osin

e.

60°

= A

U

se a

calc

ula

tor.

Exerc

ises

Fin

d x

. R

ou

nd

an

gle

me

asu

re

s t

o t

he

ne

are

st

de

gre

e a

nd

sid

e m

ea

su

re

s t

o t

he

ne

are

st

ten

th.

1

.

2.

3

.

4

.

5.

6

.

8-6

Exam

ple

1

Exam

ple

2

48

°10

12

c

C

BA

12

14

x

62°

12

11

10

x°

16

24

18x°

25

20

x

82°

18

28

x

59°

18

15

15

x°

8

5

7

C

B

A

13.5

51

42

29.8

24.3

53


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

38

G

len

co

e G

eo

me

try

Fin

d x

. R

ou

nd

an

gle

me

asu

re

s t

o t

he

ne

are

st

de

gre

e a

nd

sid

e l

en

gth

s t

o t

he

ne

are

st

ten

th.

1.

28

35°

48°

x

2.

18

46°

17°

x

3.

38

x

86°

51°

4

.

5.

6

.

7

.

8.

9

.

10

.

11

.

12

.

13

.

14

.

15

.

16

. S

olv

e t

he t

rian

gle

. R

ou

nd

an

gle

measu

res

to t

he n

eare

st d

egre

e.

Sk

ills

Practi

ce

Th

e L

aw

of

Sin

es a

nd

Law

of

Co

sin

es

8-6

x8

73°80°

x

34

41°

88°

x

12

83°

60°

18

x

104°

22°

x27

54°

93°

12

16

11

x°

17

18

23

29

111°

34x

16

89°

12x

13

103°

20

x

13

35

x

96°

28 x°

26

30

25

28°

x20

21.6

7.3

48.8

46.6

40.1

88

52

19.8

26.2

16.8

19.8

43.7

38.6

11.9

67

m∠

A =

82,

m∠

B =

51,

m∠

C =

47

Lesson 8-6


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

39

G

len

co

e G

eo

me

try

Practi

ce

Th

e L

aw

of

Sin

es a

nd

Law

of

Co

sin

es

13. IN

DIR

EC

T M

EA

SU

RE

ME

NT

T

o f

ind

th

e d

ista

nce

fro

m t

he e

dge

of

the l

ak

e t

o t

he t

ree o

n t

he i

slan

d i

n t

he l

ak

e,

Han

nah

set

up

a t

rian

gu

lar

con

figu

rati

on

as

show

n i

n t

he d

iagra

m.

Th

e d

ista

nce

from

loca

tion

A t

o l

oca

tion

B i

s 85 m

ete

rs.

Th

e m

easu

res

of

the

an

gle

s at A

an

d B

are

51°

an

d 8

3°,

resp

ect

ively

. W

hat

is t

he

dis

tan

ce f

rom

th

e e

dge o

f th

e l

ak

e a

t B

to t

he t

ree o

n t

he i

slan

d a

t C

?

A

C

B

Fin

d x

. R

ou

nd

an

gle

me

asu

re

s t

o t

he

ne

are

st

de

gre

e a

nd

sid

e l

en

gth

s t

o t

he

ne

are

st

ten

th.

1

.

2.

3

.

4

.

5.

6

.

7

.

8.

9

.

10

.

11

.

12

.

14

x

67°

14°

127

x

42°

61°

x

5.8

83°

73°

19.1

x

56°

34°

9.6

x

43° 123°

4.3

x

85°

64°

40

12

37°

x23

28

37°

x

15

17

62°

x

28.4

14.5

21.7

x°

89°

14

15

x

14

10

9.6

x°

8-6

3.7

166,

014.2

31.3

16.9

16.6

48

20.3

43

12.9

8.3

33.2

ab

ou

t 91.8

m


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

40

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

Th

e L

aw

of

Sin

es a

nd

Law

of

Co

sin

es

1

. A

LT

ITU

DE

S In

tri

an

gle

AB

C,

the

alt

itu

de t

o s

ide −

−

AB

is

dra

wn

.

A

ab

B

C

G

ive t

wo e

xp

ress

ion

s fo

r th

e l

en

gth

of

the a

ltit

ud

e i

n t

erm

s of

a,

b,

an

d t

he

sin

e o

f th

e a

ngle

s A

an

d B

.

2

. M

AP

S T

hre

e c

itie

s fo

rm t

he v

ert

ices

of

a t

rian

gle

. T

he a

ngle

s of

the t

rian

gle

are

40°,

60°,

an

d 8

0°.

Th

e t

wo m

ost

dis

tan

t ci

ties

are

40 m

iles

ap

art

. H

ow

clo

se a

re

the t

wo c

lose

st c

itie

s? R

ou

nd

you

r an

swer

to t

he n

eare

st t

en

th o

f a m

ile.

3

. S

TA

TU

ES

G

ail

was

vis

itin

g a

n a

rt

gall

ery

. In

on

e r

oom

, sh

e s

tood

so t

hat

she h

ad

a v

iew

of

two s

tatu

es,

on

e o

f a

man

, an

d t

he o

ther

of

a w

om

an

. S

he w

as

40 f

eet

from

th

e s

tatu

e o

f th

e w

om

an

, an

d 3

5 f

eet

from

th

e s

tatu

e o

f th

e m

an

. T

he a

ngle

cre

ate

d b

y t

he l

ines

of

sigh

t to

th

e t

wo s

tatu

es

was

21°.

Wh

at

is t

he

dis

tan

ce b

etw

een

th

e t

wo s

tatu

es?

R

ou

nd

you

r an

swer

to t

he n

eare

st t

en

th.

4

. C

AR

S T

wo c

ars

sta

rt m

ovin

g f

rom

th

e

sam

e l

oca

tion

. T

hey h

ead

str

aig

ht,

bu

t in

dif

fere

nt

dir

ect

ion

s. T

he a

ngle

betw

een

wh

ere

th

ey a

re h

ead

ing i

s 43

°.

Th

e f

irst

car

travels

20 m

iles

an

d t

he

seco

nd

car

travels

37 m

iles.

How

far

ap

art

are

th

e t

wo c

ars

? R

ou

nd

you

r an

swer

to t

he n

eare

st t

en

th.

5

. IS

LA

ND

S O

ah

u

is a

Haw

aii

an

Is

lan

d.

Off

of

the

coast

of

Oah

u,

there

is

a v

ery

ti

ny i

slan

d

kn

ow

n a

s C

hin

am

an

’s H

at.

K

eok

i an

d M

ali

a

are

obse

rvin

g

Ch

inam

an

’s H

at

from

loca

tion

s 5 k

ilom

ete

rs a

part

. U

se t

he i

nfo

rmati

on

in

th

e f

igu

re t

o a

nsw

er

the f

oll

ow

ing

qu

est

ion

s.

a.

How

far

is K

eok

i fr

om

Ch

inam

an

’s

Hat?

Rou

nd

you

r an

swer

to t

he

neare

st t

en

th o

f a k

ilom

ete

r.

b.

How

far

is M

ali

a f

rom

Ch

inam

an

’s

Hat?

Rou

nd

you

r an

swer

to t

he

neare

st t

en

th o

f a k

ilom

ete

r.

Ka’a

’aw

a

Keoki

Chin

am

an’s

Hat

5 k

m

Malia

Kahalu

’u

22.3

°

49.5

°

8-6

Sta

tue o

f

a m

an

Sta

tue o

f a w

om

an

35 f

t

40 f

t

21˚

a s

in B

= b

sin

A =

len

gth

of

alt

itu

de

26.1

mi

14.5

ft

26.2

mi

4.0

km

2.0

km

Lesson 8-6


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

41

G

len

co

e G

eo

me

try

En

ric

hm

en

t

Iden

titi

es

An

id

en

tity

is

an

equ

ati

on

th

at

is t

rue f

or

all

valu

es

of

the

vari

able

for

wh

ich

both

sid

es

are

defi

ned

. O

ne w

ay t

o v

eri

fy

an

id

en

tity

is

to u

se a

rig

ht

tria

ngle

an

d t

he d

efi

nit

ion

s fo

r tr

igon

om

etr

ic f

un

ctio

ns.

V

erif

y t

ha

t (s

in A

)2 +

(co

s A

)2 =

1 i

s a

n i

de

nti

ty.

(si

n A

)2 +

(co

s A

)2 =

( a

−

c ) 2

+ ( b

−

c ) 2

=

a2+

b2

−

c2

=

c2

−

c2

= 1

To c

heck

wh

eth

er

an

equ

ati

on

ma

y b

e a

n i

den

tity

, you

can

test

se

vera

l valu

es.

How

ever,

sin

ce y

ou

can

not

test

all

valu

es,

you

ca

nn

ot

be c

erta

in t

hat

the e

qu

ati

on

is

an

id

en

tity

.

T

est

sin

2x =

2 s

in x

co

s x

to

se

e i

f it

co

uld

be

an

id

en

tity

.

Try

x =

20.

Use

a c

alc

ula

tor

to e

valu

ate

each

exp

ress

ion

.

sin

2x =

sin

40

2 s

in x

cos

x =

2 (

sin

20)(

cos

20)

≈

0.6

43

≈ 2

(0.3

42)(

0.9

40)

≈

0.6

43

Sin

ce t

he l

eft

an

d r

igh

t si

des

seem

equ

al,

th

e e

qu

ati

on

may b

e a

n i

den

tity

.

Exerc

ises

Use

tria

ng

le ABC

sh

ow

n a

bo

ve

. V

erif

y t

ha

t e

ach

eq

ua

tio

n i

s a

n i

de

nti

ty.

1

. co

s A

−

sin

A =

1

−

tan

A

2. ta

n B

−

sin

B =

1

−

cos

B

3. ta

n B

cos

B =

sin

B

4. 1 -

(co

s B

)2 =

(si

n B

)2

Try

se

ve

ra

l v

alu

es f

or x

to

te

st

wh

eth

er e

ach

eq

ua

tio

n c

ou

ld b

e a

n i

de

nti

ty.

5

. co

s 2x =

(co

s x)2

- (

sin

x)2

6. co

s (9

0 -

x)

= s

in x

B

AC

ca

b

8-6

Exam

ple

1

Exam

ple

2

tan

B

−

sin

B =

b

−

a

÷ b

−

c =

c

−

a =

1

−

co

s B

c

os

A

−

sin

A =

b

−

c

÷ a

−

c =

b

−

a =

1

−

tan

A

tan

B c

os B

= b

−

a

· a

−

c =

b

−

c =

sin

B1 -

(co

s B

)2 =

1 -

( a

−

c ) 2

=

c2

−

c

2 -

a

2

−

c

2

=

c2 -

a2

−

c

2

= b

2

−

c

2 o

r (s

in B

)2

Yes;

see s

tud

en

ts’

wo

rk.

Yes;

see s

tud

en

ts’

wo

rk.


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

42

G

len

co

e G

eo

me

try

Gra

ph

ing C

alc

ula

tor

Act

ivit

y

So

lvin

g T

rian

gle

s U

sin

g t

he L

aw

of

Sin

es o

r C

osin

es

You

can

use

a c

alc

ula

tor

to s

olv

e t

rian

gle

s u

sin

g t

he L

aw

of

Sin

es

or

Cosi

nes.

S

olv

e △ABC i

f a

= 6

, b

= 2

, a

nd

c =

7.5

.

U

se t

he L

aw

of

Cosi

nes.

a

2 =

b2 +

c2 -

2bc c

os A

6

2 =

22 +

7.5

2 -

2(2

)(7.5

) co

s A

m

∠A =

cos-

1 6

2 -

22 -

7.5

2

−

-2(2

)(7.5

)

Use

you

r ca

lcu

lato

r to

fin

d t

he m

easu

re o

f ∠A.

K

ey

str

ok

es:

2nd

[C

OS

-1]

( 6

x

2

— 2

x

2

— 7

.5

x2

)

÷

En

ter:

(

(–) 2

3

2

3 7

.5

)

)

EN

TE

R

36.06658826

S

o m

∠A

≈ 3

6.

Use

th

e L

aw

of

Sin

es

an

d y

ou

r ca

lcu

lato

r to

fin

d m

∠B.

si

n A

−

a =

sin

B

−

b

si

n 3

6

−

6

≈

si

n B

−

2

m

∠B ≈

sin

-1 2

sin

36°

−

6

K

ey

str

ok

es:

2nd

[S

IN-

1]

( 2

S

IN 3

6

)

÷ 6

)

E

NT

ER

11.29896425

So m

∠B

≈ 1

1.

By t

he T

rian

gle

An

gle

-Su

m T

heore

m, m

∠C ≈

180 -

(36 +

11)

or

133.

Exerc

ises

So

lve

ea

ch

△ABC

. R

ou

nd

me

asu

re

s o

f sid

es t

o t

he

ne

are

st

ten

th

an

d m

ea

su

re

s o

f a

ng

les t

o t

he

ne

are

st

de

gre

e.

1

. a

= 9

, b

= 1

4, c =

12

2

. m

∠C

= 8

0, c =

9, m

∠A

= 4

0

3

. m

∠B

= 4

5, m

∠C

= 5

6, a

= 2

4

. a =

5.7

, b =

6, c =

5

5

. a =

11, b =

15, c =

21

8-6 Exam

ple

m∠

A ≈

40,

m∠

B ≈

89,

m∠

C ≈

51

m∠

B =

60,

a ≈

5.9

, b

≈ 7

.9

m∠

A =

79,

b ≈

1.4

, c ≈

1.7

m∠

A =

62,

m∠

B =

68,

m∠

C =

50

m∠

A ≈

30,

m∠

B ≈

43,

m∠

C ≈

107

Lesson 8-7


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

43

G

len

co

e G

eo

me

try

Geo

metr

ic V

ect

or

Op

era

tio

ns

A v

ect

or

is a

dir

ect

ed s

egm

en

t re

pre

sen

tin

g a

qu

an

tity

th

at

has

both

ma

gn

itu

de, or

len

gth

, an

d d

irecti

on

. F

or

exam

ple

, th

e s

peed a

nd d

irect

ion

of

an

air

pla

ne c

an

be r

epre

sen

ted b

y a

vect

or.

In

sym

bols

, a v

ect

or

is w

ritt

en

as

'

AB ,

wh

ere

A i

s th

e

init

ial

poin

t an

d B

is

the e

ndpoin

t, o

r as

'

v .

Th

e s

um

of

two v

ect

ors

is

call

ed t

he r

esu

lta

nt.

S

ubtr

act

ing a

vect

or

is e

qu

ivale

nt

to a

ddin

g i

ts o

pposi

te. T

he r

esu

ltan

t of

two v

ect

ors

can

be

fou

nd u

sin

g t

he p

ara

llelo

gra

m m

eth

od

or

the t

ria

ng

le m

eth

od

.

C

op

y t

he

ve

cto

rs t

o f

ind

$ a -

$ b .

a(

b(

Me

tho

d 1

: U

se t

he p

ara

llelo

gra

m m

eth

od

.

Co

py ' a

an

d -

'

b w

ith

th

e s

am

e in

itia

l

po

int.

a

-b

Co

mp

lete

th

e p

ara

llelo

gra

m.

a

-b

Dra

w t

he

dia

go

na

l o

f th

e

pa

ralle

log

ram

fro

m t

he

in

itia

l p

oin

t.

a

a

-b

-b

Me

tho

d 2

: U

se t

he t

rian

gle

meth

od

.

Co

py ' a

.

a

Pla

ce

th

e in

itia

l p

oin

t o

f -

'

b a

t th

e

term

ina

l p

oin

t o

f ' a .

a-b

Dra

w t

he

ve

cto

r fr

om

th

e in

itia

l

po

int

of

'

a t

o t

he

te

rmin

al p

oin

t o

f

- '

b

.

a -b

a -

b

Exerc

ises

Co

py

th

e v

ecto

rs.

Th

en

fin

d e

ach

su

m o

r d

iffe

re

nce

.

1

. ' c

+ '

d

2. w

- '

z

3. ' a

- '

b

4. ' r

+ '

t

((

Exam

ple

c(d(

c(d(

+c(

d(

w(

z(

w(

w(z(

-( z

r(

t(

r(t(

+

r(

t(

a(

b(

a( -b(

a(

b(

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Vecto

rs

8-7

Answers (Lesson 8-6 and Lesson 8-7)

Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

44

G

len

co

e G

eo

me

try

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(conti

nued

)

Vectors

8-7

Ve

cto

rs o

n t

he

Co

ord

ina

te

Pla

ne

A

vect

or

in s

tan

da

rd

po

sit

ion

h

as

its

init

ial

poin

t at

(0,

0)

an

d c

an

be r

ep

rese

nte

d b

y t

he o

rdere

d

pair

for

poin

t B

. T

he v

ect

or

at

the r

igh

t ca

n b

e e

xp

ress

ed

as

� v =

⟨5,

3⟩.

You

can

use

th

e D

ista

nce

Form

ula

to f

ind

th

e m

agn

itu

de

| �

AB

| of

a v

ect

or.

You

can

desc

ribe t

he d

irect

ion

of

a v

ect

or

by m

easu

rin

g t

he a

ngle

th

at

the v

ect

or

form

s w

ith

th

e p

osi

tive

x-a

xis

or

wit

h a

ny o

ther

hori

zon

tal

lin

e.

F

ind

th

e m

ag

nit

ud

e a

nd

dir

ecti

on

of

a =

⟨3

, 5

⟩.

Fin

d t

he m

agn

itu

de.

� a

= √

%%

%%

%%

%%

%

(x

2 -

x1)2

+ (y

2 -

y1)2

=

√ %

%%

%%

%%

%

(3

- 0

)2 +

(5

- 0

)2

=

√ %%

34 o

r abou

t 5.8

To f

ind

th

e d

irect

ion

, u

se t

he t

an

gen

t ra

tio.

ta

n θ

= 5

−

3

The t

angent

ratio is o

pposite o

ver

adja

cent.

m

∠θ ≈

59.0

U

se a

calc

ula

tor.

Th

e m

agn

itu

de o

f th

e v

ect

or

is a

bou

t 5.8

un

its

an

d i

ts d

irect

ion

is

59°.

Exerc

ises

Fin

d t

he

ma

gn

itu

de

an

d d

ire

cti

on

of

ea

ch

ve

cto

r.

1. � b

= ⟨

-5, 2

⟩ 2.

� c =

⟨-

2, 1

⟩

3. � d =

⟨3, 4

⟩ 4.

� m =

⟨5,

-1

⟩

5. � r

= ⟨

-3,

-4

⟩ 6.

� v =

⟨-

4, 1

⟩

x

y

( 3, 5)

x

y

( 5, 3)

( 0, 0)

Exam

ple

5.4

; 158.2

°2.2

; 153.4

°

5;

53.1

°5.1

; 348.7

°

5;

233.1

°4.1

; 166.0

°

Dis

tance F

orm

ula

(x1,

y1)

= (

0,

0)

and (

x2,

y2)

= (

3,

5)

Sim

plif

y.

Lesson 8-7


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

45

G

len

co

e G

eo

me

try

Sk

ills

Pra

ctic

e

Vectors

8-7

Use

a r

ule

r a

nd

a p

ro

tra

cto

r t

o d

ra

w e

ach

ve

cto

r.

Inclu

de

a s

ca

le o

n e

ach

dia

gra

m.

1.

� a =

20 m

ete

rs p

er

seco

nd

2.

� b =

10 p

ou

nd

of

forc

e a

t

60°

west

of

sou

th

135°

to t

he h

ori

zon

tal

Co

py

th

e v

ecto

rs t

o f

ind

ea

ch

su

m o

r d

iffe

re

nce

.

3

. � a

+ � z

4.

� t -

� r

Writ

e t

he

co

mp

on

en

t fo

rm

of

ea

ch

ve

cto

r.

5

.

6.

Fin

d t

he

ma

gn

itu

de

an

d d

ire

cti

on

of

ea

ch

ve

cto

r.

7.

� m =

⟨2,

12

⟩ 8

. � p

= ⟨

3,

10

⟩

9.

� k =

⟨-

8,

-3

⟩ 1

0.

� f =

⟨-

5,

11

⟩

Fin

d e

ach

of

the

fo

llo

win

g f

or

a

= ⟨

2,

4⟩,

b =

⟨3

, -

3⟩

, a

nd

c =

⟨4

, -

1⟩.

Ch

eck

yo

ur

an

sw

ers g

ra

ph

ica

lly

.

x

y

O

B( –2, 2)

D( 3, –3)

x

y

O

C( 1, –1)

E( 4, 3)

a

z

r t

-r

r-

t

t a

z

az

+

N

E

S

W

1 cm : 5 m

60°

a

1 in : 10 lb135°

b

Overm

att

er

2 √

''

37 ≈

12.2

, 80.5

° √

''

109 ≈

10.4

, 73.3

°

√ ''

73 ≈

8.5

, 200.6

° √

''

146 ≈

12.1

, 155.6

°

⟨3,

4⟩

⟨5,

-5

⟩


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

46

G

len

co

e G

eo

me

try

Practi

ce

Vectors

Use

a r

ule

r a

nd

a p

ro

tra

cto

r t

o d

ra

w e

ach

ve

cto

r.

Inclu

de

a s

ca

le o

n e

ach

dia

gra

m.

1.

� v =

12 N

ew

ton

s of

forc

e a

t 2. � w =

15 m

iles

per

hou

r

40°

to t

he h

ori

zon

tal

70°

east

of

nort

h

Co

py

th

e v

ecto

rs t

o f

ind

ea

ch

su

m o

r d

iffe

re

nce

.

3

. � p

+ � r

4. � a

- � b

5. W

rite

th

e c

om

pon

en

t fo

rm o

f

�

AB .

Fin

d t

he

ma

gn

itu

de

an

d d

ire

cti

on

of

ea

ch

ve

cto

r.

6

. � t

= ⟨

6,

11

⟩ 7

. � g

= ⟨

9,

-7

⟩

Fin

d e

ach

of

the

fo

llo

win

g f

or

a

= ⟨

-1

.5,

4⟩,

b =

⟨7

, 3

⟩, a

nd

c =

⟨1

, -

2⟩.

Ch

eck

yo

ur

an

sw

ers g

ra

ph

ica

lly

.

8.

2 � a

+ � b

9. 2 � c

- � b

10.

AV

IAT

ION

A

jet

begin

s a f

ligh

t alo

ng a

path

du

e n

ort

h a

t 300 m

iles

per

hou

r. A

win

d

is b

low

ing d

ue w

est

at

30 m

iles

per

hou

r.

a.

Fin

d t

he r

esu

ltan

t velo

city

of

the p

lan

e.

b.

Fin

d t

he r

esu

ltan

t d

irect

ion

of

the p

lan

e.

8-7

x

y

O

K( –2, 4)

L( 3, –4)

p%r%

p

r p

+r

a%b%

-b

a-

b

a

1 cm : 4 N

40°

V

N

ES

W

1 in : 10 m

i

70°

w

⟨5,

-8

⟩

ab

ou

t 301.5

mp

h

ab

ou

t 5.7

° west

of

du

e n

ort

h

b 2

a

x

y

−448

12

4−4

812

2a

+ b

b

y

−4

−848

x

−8

−12

4

2c

-b

2c

⟨-1,

13

⟩⟨-

5,

-1

⟩

Lesson 8-7


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

47

G

len

co

e G

eo

me

try

Wo

rd

Pro

ble

m P

racti

ce

Vectors

8-7

1

. W

IND

T

he v

ect

or

�v re

pre

sen

ts t

he s

peed

an

d d

irect

ion

th

at

the w

ind

is

blo

win

g.

Su

dd

en

ly t

he w

ind

pic

ks

up

an

d d

ou

ble

s

its

speed

, bu

t th

e d

irect

ion

does

not

chan

ge.

Wri

te a

n e

xp

ress

ion

for

a v

ect

or

that

desc

ribes

the n

ew

win

d v

elo

city

in

term

s of

�v.

2

. SW

IMM

ING

Jan

is

swim

min

g i

n a

tria

thlo

n e

ven

t. W

hen

th

e o

cean

wate

r

is s

till

, h

er

velo

city

can

be r

ep

rese

nte

d

by t

he v

ect

or

⟨2,

1⟩

mil

es

per

hou

r.

Du

rin

g t

he c

om

peti

tion

, th

ere

was

a

fierc

e c

urr

en

t re

pre

sen

ted

by t

he

vect

or

⟨–1,

–1

⟩ m

iles

per

hou

r. W

hat

vect

or

rep

rese

nts

Jan

’s v

elo

city

du

rin

g

the r

ace

?

3

. PO

LY

GO

NS

D

raw

a r

egu

lar

poly

gon

aro

un

d t

he o

rigin

. F

or

each

sid

e o

f th

e

poly

gon

, ass

oci

ate

a v

ect

or

wh

ose

magn

itu

de i

s th

e l

en

gth

of

the

corr

esp

on

din

g s

ide a

nd

wh

ose

dir

ect

ion

poin

ts i

n t

he c

lock

wis

e m

oti

on

aro

un

d

the o

rigin

. W

hat

vect

or

rep

rese

nts

th

e

sum

of

all

th

ese

vect

ors

? E

xp

lain

.

4

. B

AS

EB

ALL R

ick

is

in t

he m

idd

le o

f a

base

ball

gam

e.

His

team

mate

th

row

s

him

th

e b

all

, bu

t th

row

s it

far

in f

ron

t of

him

. H

e h

as

to r

un

as

fast

as

he c

an

to

catc

h i

t. A

s h

e r

un

s, h

e k

now

s th

at

as

soon

as

he c

atc

hes

it,

he h

as

to t

hro

w i

t

as

hard

as

he c

an

to t

he t

eam

mate

at

hom

e p

late

. H

e h

as

no t

ime t

o s

top

. In

the f

igu

re,

�x is

th

e v

ect

or

that

rep

rese

nts

the v

elo

city

of

the b

all

after

Ric

k t

hro

ws

it a

nd

�v

rep

rese

nts

Ric

k’s

velo

city

beca

use

he i

s ru

nn

ing.

Ass

um

e t

hat

Ric

k

can

th

row

ju

st a

s h

ard

wh

en

ru

nn

ing a

s

he c

an

wh

en

sta

nd

ing s

till

. Homeplate

Rick

x

v%

%% X

% V

a.

Wh

at

vect

or

wou

ld r

ep

rese

nt

the

velo

city

of

the b

all

if

Ric

k t

hre

w i

t th

e

sam

e w

ay b

ut

he w

as

stan

din

g s

till

?

b.

Th

e a

ngle

betw

een

� x an

d � v

is 8

9°.

By

run

nin

g,

did

it

help

Ric

k g

et

the b

all

to h

om

e p

late

fast

er

than

he w

ou

ld

have n

orm

all

y b

een

able

to i

f h

e w

ere

stan

din

g s

till

?

2

v

⟨1,

0⟩

mp

h

x -

v

⟨0,

0⟩;

Becau

se t

he p

oly

go

n

form

s a

clo

sed

lo

op

, th

e s

um

is

th

e z

ero

vecto

r.

Yes,

it h

elp

ed

.


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

8

48

G

len

co

e G

eo

me

try

En

ric

hm

en

t8-7

Do

t P

ro

du

ct

Th

e d

ot

pro

du

ct o

f tw

o v

ect

ors

rep

rese

nts

how

mu

ch t

he v

ect

ors

poin

t in

th

e d

irect

ion

of

each

oth

er.

If

� v is

a v

ect

or

rep

rese

nte

d b

y ⟨a

, b

⟩ an

d � u

is a

vect

or

rep

rese

nte

d b

y ⟨c, d

⟩, t

he

form

ula

to f

ind

th

e d

ot

pro

du

ct i

s:

� v ·

� u = ac +

bd

Look

at

the f

oll

ow

ing e

xam

ple

:

Gra

ph

th

e v

ect

ors

an

d f

ind

th

e d

ot

pro

du

ct o

f � v

an

d � u

if

� v =

⟨3,

-1

⟩ an

d � u

= ⟨

2,

5⟩.

� v ·

� u =

(3)(

2)

+ (

-1)(

5)

or

1

Gra

ph

th

e v

ecto

rs a

nd

fin

d t

he

do

t p

ro

du

cts

.

1

. � v

= ⟨

2,

1⟩

an

d � u

= ⟨

-4,

2⟩

2. � v

= ⟨

3,

-2

⟩ an

d � u

= ⟨

1,

4⟩

Th

e d

ot

pro

du

ct i

s

T

he d

ot

pro

du

ct i

s

3. � v

= ⟨

0,

3⟩

an

d � u

= ⟨

2,

4⟩

4. � v

= ⟨

-1,

4⟩

an

d � u

= ⟨

-4,

2⟩

Th

e d

ot

pro

du

ct i

s

T

he d

ot

pro

du

ct i

s

5. N

oti

ce t

he a

ngle

form

ed

by t

he t

wo v

ect

ors

an

d t

he c

orr

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Chapter 8 Assessment Answer KeyQuiz 1 (Lessons 8-1 and 8-2) Quiz 3 (Lessons 8-5 and 8-6) Mid-Chapter Test

Page 51 Page 52 Page 53

Quiz 2 (Lessons 8-3 and 8-4)

Page 51

Quiz 4 (Lesson 8-7)

Page 52

Part I

Part II

1.

2.

3.

4.

5.

8 √ " 3

x = √ "" 85 ,

y = 2 √ "" 15

x = √ "" 17 , y = 81

− 8

√ "" 137

acute

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

3 √ " 2

2 √ " 3

57.8

14.6

20 √ " 3 + 20 in.

0.7880

27

3.3 ft

27.1

22.0

1.

2.

3.

4.

5.

∠QPR

3.9

m∠B = 104,m∠C = 27,b = 23.2

m∠R = 88,m∠T = 53,m∠S = 39

52°

1.

2.

3.

4.

5.

24.8 units, 220°

31.4 units, 158°

6 mph, south

c"d"

c" d"+

k" k"

-m"

-m"

6.

7.

8.

9.

10.

x = 9, y = 3 √ " 3

x = 10 √ " 3 ,

y = 10 √ " 6

x = 12 √ " 2 ,

y = 24 √ " 2

right

54

1.

2.

3.

4.

5.

A

H

B

H

D

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Chapter 8 Assessment Answer KeyVocabulary Test Form 1

Page 54 Page 55 Page 56

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

geometric mean

Pythagorean triple

false,trigonometric

ratio

true

angle of depression

trigonometry

sine

tangent

A method to describe a vector using its

horizontal and vertical change from its inital

to terminal point.

In a right triangle, the sum of the squares of

the measures of the

legs equals the square

of the measure of the

hypotenuse.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

D

F

A

G

D

G

C

G

C

H

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

C

F

A

G

D

F

B

F

C

G

69

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Answers


Chapter 8 Assessment Answer KeyForm 2A Form 2B

Page 57 Page 58 Page 59 Page 60

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

D

J

A

H

D

J

A

G

B

H

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

B

F

C

G

D

H

C

G

D

H

56.4 ft

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

A

J

D

G

D

F

A

H

B

H

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

B

H

C

G

D

G

A

G

B

J

73.2 ft

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Chapter 8 Assessment Answer KeyForm 2C

Page 61 Page 62

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

√ "" 10

13

2 √ " 7

√ """ 4000 or 20 √ "" 10

√ "" 300 or 10 √ " 3

11 √ " 2

2 √ " 3

x = 6, y = 12

9.7

69

68

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

267.9m

11°

43.2

14.3

15.6 ft

20

33.0

√ "" 170 ≈ 13 units; 237.5°

1

s

t

s t+

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Chapter 8 Assessment Answer KeyForm 2D

Page 63 Page 64

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

3 √ "" 10

26

√ "" 39

√ """ 7300 or 10 √ "" 73

√ "" 432 or 12 √ " 3

15 √ " 2

7 √ " 3

x = 12, y = 24

8.6

70°

67°

352.7 m

9°

32.3

6.2°

15.2 ft

19

38.2

√ "" 109 ≈ 10.4 units, 253.3°

6 or 10

d

-f

d f-

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

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Chapter 8 Assessment Answer KeyForm 3

Page 65 Page 66

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

√ " 6

− 9

2

3

2 √ "" 10

12 −

5

12.5

250 km

right

12 √ " 6

24 + 8 √ " 3

x = 5 √ " 3 ,

y = 10 √ " 3

(-4 + 8 √ " 3 , -2)

or (-4 - 8 √ " 3 , -2)

13.

14.

15.

16.

17.

18.

19.

20.

B:

x = 5 √ " 6

− 3 , CD = 5

√ " 6 −

3

+ 5 √ " 2 + 12

147 ft

2 ft

37.4

253.7 yd

37° or 143°

31.4 units, 261°

71.1 ft and 84.5 ft

a

ab

-b

a b-

a b+

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Chapter 8 Assessment Answer KeyExtended-Response Test, Page 67

Scoring Rubric

Score General Description Specific Criteria

4 Superior

A correct solution that

is supported by well-

developed, accurate

explanations

• Shows thorough understanding of the concepts of geometric mean,

special right triangles, altitude to the hypotenuse theorems,

Pythagorean Theorem, solving triangles, SOH CAH TOA, Law of

Sines, and Law of Cosines.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• Figures are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution,

but may contain minor flaws

in reasoning or computation

• Shows an understanding of the concepts of geometric mean,

special right triangles, altitude to the hypotenuse theorems,


Sines, and Law of Cosines.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Figures are mostly accurate and appropriate.

• Satisfies all requirements of problems.

2 Nearly Satisfactory

A partially correct

interpretation and/or

solution to the problem

• Shows an understanding of most of the concepts of geometric

mean, special right triangles, altitude to the hypotenuse theorems,


Sines and Law of Cosines.

• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Figures are mostly accurate.

• Satisfies the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no

supporting evidence or

explanation

• Final computation is correct.

• No written explanations or work is shown to substantiate the final

computation.

• Figures may be accurate but lack detail or explanation.

• Satisfies minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution

indicating no mathematical

understanding of the

concept or task, or no

solution is given

• Shows little or no understanding of most of the concepts of

geometric mean, special right triangles, altitude to the hypotenuse

theorems, Pythagorean Theorem, solving triangles, SOH CAH

TOA, Law of Sines and Law of Cosines.

• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Figures are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer given.

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Chapter 8 Assessment Answer KeyExtended-Response Test, Page 67

Sample Answers

1. 10

− 6 =

6 −

x

10x = 36

x = 18

− 5

2a. No, his work is not correct. The triangle is not a right triangle so he cannot use the altitude to the hypotenuse theorem. Instead he must use the 30°-60°-90° triangle relationships and obtain

x = 2 √ $ 3 .

b. Using the 30°-60°-90° triangle relationships, RQ = 4, and therefore PQ would have to equal 8, so PS = 8 - 2 or 6.

c. No, the sides are not in a ratio of 1:1: √ $ 2 .

3. If you were finding the length of a side x you would use sin of the given angle. If you are finding an angle x you would use the s in -1 of the ratio of two sides to find the angle. For example, if

sin x = 3 −

4 , use s in -1

3 −

4 to find x. If

sin 20° = x −

5 find the sine of 20° and

multiply by 5 to find x, the length of the unknown side.

4. angle of depression

angle of elevation

Student should draw a right triangle and label the angle of elevation up from the horizontal and the angle of depression down from the horizontal. The angle of elevation has the same measure as the angle of depression.

5.

150A C

B

20°

110°ca

Let m∠B = 110, m∠A = 20, and b = 150. Use the Law of Sines to find the missing lengths. The length of a is found using

sin 110°

− 150

= sin 20°

− a . The third angle is

found by evaluating 180 - (m∠B + m∠A). In this problem, m∠C = 50. The length of

c is found by using sin 110°

− 150

= sin 50°

− c .

In this problem, c ≈ 122.3.

6. No, her plan is not a good one. Irina should use the Law of Cosines to find angle B. Then she can use the Law of Sines to find either angle A or angle C and subtract the two angles she finds from 180 degrees to get the measure of the third angle.

In addition to the scoring rubric found on page A31, the following sample answers may be used as guidance in evaluating open-ended assessment items.

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Answers


Chapter 8 Assessment Answer KeyStandardized Test Practice

Page 68 Page 69

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

15.

16.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

1 8 0.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

1 4 5.

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

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Chapter 8 Assessment Answer KeyStandardized Practice Test

Page 70

17.

18.

19.

20.

21.

22.

23a.

23b.

23c.

Def. of segments

HJ + JK = HK;KL + LM = KM

m∠1 = 32, m∠2 = 82,

m∠3 = 66, m∠4 =

114, m∠5 = 57

7

19

∠3, ∠4, ∠5, ∠7,∠ABC

yes, both are 30°- 60°- 90°

triangles; AASimilarity

36

2:1

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Chapter 8 Resource Masters - · PDF fileThe Chapter 8 Resource Masters includes the core materials needed for Chapter ... students on how to assess performance on ... Mid-Chapter Quiz