Teacher ToolboxPRO 2 - in Trigonometry 6v2.toolboxpro.org/secure/teachers/1162/pc6_2e.pdf · 2012-06-11 · oblique triangles (AAS or ASA). • Use the Law of Sines to solve oblique

6Additional Topics in Trigonometry

6.1 Law of Sines

6.2 Law of Cosines

6.3 Vectors in the Plane

6.4 Vectors and Dot Products

6.5 Trigonometric Form of a Complex Number

In Mathematics

Trigonometry is used to solve triangles,

represent vectors, and to write trigonometric

forms of complex numbers.

In Real Life

Trigonometry is used to find areas, estimate

heights, and represent vectors involving

force, velocity, and other quantities.

For instance, trigonometry and vectors can

be used to find the tension in the tow lines

as a loaded barge is being towed by two

tugboats. (See Exercise 93, page 456.)

IN CAREERS

There are many careers that use trigonometry. Several are listed below.

Luca

Tettoni/Terra/C

orbis

• Pilot

Exercise 51, page 435

• Civil Engineer


• Awning Designer


• Landscaper


427

428 Chapter 6 Additional Topics in Trigonometry

6.1 LAW OF SINES

What you should learn

• Use the Law of Sines to solveoblique triangles (AAS or ASA).

• Use the Law of Sines to solveoblique triangles (SSA).

• Find the areas of oblique triangles.

• Use the Law of Sines to model andsolve real-life problems.

Why you should learn it

You can use the Law of Sines to solvereal-life problems involving obliquetriangles. For instance, in Exercise 53on page 436, you can use the Law ofSines to determine the distance froma boat to the shoreline.

© O

wen

Frank

en/C

orbis

Introduction

In Chapter 4, you studied techniques for solving right triangles. In this section and the

next, you will solve oblique triangles—triangles that have no right angles. As standard

notation, the angles of a triangle are labeled and and their opposite sides are

labeled and as shown in Figure 6.1.

FIGURE 6.1

To solve an oblique triangle, you need to know the measure of at least one side and

any two other measures of the triangle—either two sides, two angles, or one angle and

one side. This breaks down into the following four cases.

1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

The first two cases can be solved using the Law of Sines, whereas the last two cases

require the Law of Cosines (see Section 6.2).

The Law of Sines can also be written in the reciprocal form

For a proof of the Law of Sines, see Proofs in Mathematics on page 487.

sin A

a�

sin B

b�

sin C

c.

A B

a

c

b

C

c,b,a,

C,B,A,

Law of Sines

If is a triangle with sides and then

A is acute. A is obtuse.

A B

ab

c

h

C

A B

ab

c

h

C

a

sin A�

b

sin B�

c

sin C.

c,b,a,ABC

Section 6.1 Law of Sines 429

Given Two Angles and One Side—AAS

For the triangle in Figure 6.2, and feet. Find the remaining

angle and sides.

Solution

The third angle of the triangle is

By the Law of Sines, you have

Using produces

and

Now try Exercise 5.

Given Two Angles and One Side—ASA

A pole tilts toward the sun at an angle from the vertical, and it casts a 22-foot shadow.

The angle of elevation from the tip of the shadow to the top of the pole is How tall

is the pole?

Solution

From Figure 6.3, note that and So, the third angle is


Because feet, the length of the pole is

Now try Exercise 45.

For practice, try reworking Example 2 for a pole that tilts away from the sun under

the same conditions.

a �c

sin C�sin A� �

22

sin 39 �sin 43 � 23.84 feet.

c � 22

a

sin A�

c

sin C.

� 39 .

� 180 � 43 � 98

C � 180 � A � B

B � 90 � 8 � 98 .A � 43

43 .

8

Example 2

c �b

sin B�sin C� �

28

sin 29 �sin 102 � 56.49 feet.

a �b

sin B�sin A� �

28

sin 29 �sin 49 � 43.59 feet

b � 28

a

sin A�

b

sin B�

c

sin C.

� 49 .

� 180 � 29 � 102

A � 180 � B � C

b � 28C � 102 , B � 29 ,

Example 1

A B

a

c29°

102°

b = 28 ftC

FIGURE 6.2

AB

43°

8°a

b

c = 22 ft

C

FIGURE 6.3

When solving triangles, a careful

sketch is useful as a quick test

for the feasibility of an answer.

Remember that the longest side

lies opposite the largest angle,

and the shortest side lies opposite

the smallest angle.


The Ambiguous Case (SSA)

In Examples 1 and 2, you saw that two angles and one side determine a unique triangle.

However, if two sides and one opposite angle are given, three possible situations can

occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles

may satisfy the conditions.

Single-Solution Case—SSA

For the triangle in Figure 6.4, inches, inches, and Find the

remaining side and angles.

Solution


Reciprocal form

Multiply each side by b.

Substitute for A, a, and b.

B is acute.

Now, you can determine that

Then, the remaining side is


c �a

sin A�sin C� �

22

sin 42 �sin 116.59 � 29.40 inches.

c

sin C�

a

sin A

� 116.59 .C 180 � 42 � 21.41

B 21.41 .

sin B � 12�sin 42

22

sin B � b�sin A

a

sin B

b�

sin A

a

A � 42 .b � 12a � 22

Example 3

The Ambiguous Case (SSA)

Consider a triangle in which you are given and

is acute. is acute. is acute. is acute. is obtuse. is obtuse.

Sketch

Necessary

condition

Triangles None One One Two None One

possible

a > ba ! bh < a < ba � ba � ha < h

ab

A

a

b

A

a ah

b

AA

ab hahb

AA

b ah

AAAAAA

�h � b sin A�A.b,a,

a = 22 in.b = 12 in.

A B42°

c

C

One solution: a b

FIGURE 6.4

�


No-Solution Case—SSA

Show that there is no triangle for which and

Solution

Begin by making the sketch shown in Figure 6.5. From this figure it appears that no

triangle is formed. You can verify this using the Law of Sines.

Reciprocal form

Multiply each side by b.

This contradicts the fact that So, no triangle can be formed having sides

and and an angle of


Two-Solution Case—SSA

Find two triangles for which meters, meters, and

Solution


Reciprocal form

There are two angles, and between and

whose sine is 0.9047. For you obtain

For you obtain

The resulting triangles are shown in Figure 6.6.

FIGURE 6.6


A

a = 12 mb = 31 m

20.5°115.2°

B2A

a = 12 mb = 31 m

20.5° 64.8°B1

c �a


12

sin 20.5 �sin 44.3 � 23.93 meters.

C 180 � 20.5 � 115.2 � 44.3

B2 115.2 ,

c �a


12

sin 20.5 �sin 94.7 � 34.15 meters.

C 180 � 20.5 � 64.8 � 94.7

B1 64.8 ,180

0 B2 180 � 64.8 � 115.2 ,B1 64.8

sin B � b�sin A

a � 31�sin 20.5

12 0.9047.

sin B

b�

sin A

a

A � 20.5 .b � 31a � 12

Example 5

A � 85 .b � 25a � 15sin B ! 1.

sin B � 25�sin 85

15 1.660 > 1

sin B � b�sin A

a

sin B

b�

sin A

a

A � 85 .a � 15, b � 25,

Example 4

A

a = 15

b = 25

85°

h

No solution: a h

FIGURE 6.5

<


Area of an Oblique Triangle

The area of any triangle is one-half the product of the lengths of two sides times

the sine of their included angle. That is,

Area �1

2bc sin A �

1

2ab sin C �

1

2ac sin B.

Area of an Oblique Triangle

The procedure used to prove the Law of Sines leads to a simple formula for the area of

an oblique triangle. Referring to Figure 6.7, note that each triangle has a height of

Consequently, the area of each triangle is

By similar arguments, you can develop the formulas

A is acute. A is obtuse.

FIGURE 6.7

Note that if angle is the formula gives the area for a right triangle:

Similar results are obtained for angles and equal to

Finding the Area of a Triangular Lot

Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters

and an included angle of

Solution

Consider meters, meters, and angle as shown in Figure 6.8.

Then, the area of the triangle is


Area �1

2ab sin C �

1

2�90��52��sin 102 � 2289 square meters.

C � 102 ,b � 52a � 90

102 .

Example 6

90 .BC

sin 90 � 1Area �1

2bc�sin 90 � �

1

2bc �

1

2�base��height�.

90 ,A

AA

B

a ab b

c

h

Bc

h

C C

Area �1

2ab sin C �

1

2ac sin B.

Area �1

2�base��height� �

1

2�c��b sin A� �

1

2bc sin A.

h � b sin A.To see how to obtain the height of

the obtuse triangle in Figure 6.7,

notice the use of the reference

angle and the difference

formula for sine, as follows.

� b sin A

� b�0 " cos A � ��1� " sin A�

� cos 180 sin A�

� b�sin 180 cos A

h � b sin�180 � A�

180 � A

a = 90 m

b = 52 m

C

102°

FIGURE 6.8


Application

An Application of the Law of Sines

The course for a boat race starts at point in Figure 6.9 and proceeds in the direction

S W to point then in the direction S E to point and finally back to

Point lies 8 kilometers directly south of point Approximate the total distance of

the race course.

Solution

Because lines and are parallel, it follows that Consequently,

triangle has the measures shown in Figure 6.10. The measure of angle is

Using the Law of Sines,

Because

and

The total length of the course is approximately


� 19.453 kilometers.

Length 8 � 6.308 � 5.145

c �8

sin 88 �sin 40 � 5.145.

a �8

sin 88 �sin 52 � 6.308

b � 8,

a

sin 52 �

b

sin 88 �

c

sin 40 .

180 � 52 � 40 � 88 .

BABC

BCA ! CBD.ACBD

A.C

A.C,40 B,52

A

Example 7

Using the Law of Sines In this section, you have been using the Law of Sines tosolve oblique triangles. Can the Law of Sines also be used to solve a right triangle? If so, write a short paragraph explaining how to use the Law of Sines to solve eachtriangle. Is there an easier way to solve these triangles?

a. AAS b. ASA

50°

AC

a = 10

B

50°

AC

c = 20

B

#$#$

CLASSROOM DISCUSSION

A

B

CD

40°

52°

8 km

S

EW

N

FIGURE 6.9

40°

52°

B

C

a

c

b = 8 km

A

FIGURE 6.10


EXERCISES See www.CalcChat.com for worked-out solutions to odd-numbered exercises.6.1

VOCABULARY: Fill in the blanks.

1. An ________ triangle is a triangle that has no right angle.

2. For triangle the Law of Sines is given by

3. Two ________ and one ________ determine a unique triangle.

4. The area of an oblique triangle is given by

SKILLS AND APPLICATIONS

12bc sin A �

12ab sin C � ________ .

a

sin A� ________ �

c

sin C.ABC,

In Exercises 5–24, use the Law of Sines to solve the triangle.Round your answers to two decimal places.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

In Exercises 25–34, use the Law of Sines to solve (if possible)the triangle. If two solutions exist, find both. Round youranswers to two decimal places.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

In Exercises 35–38, find values for such that the trianglehas (a) one solution, (b) two solutions, and (c) no solution.

35.

36.

37.

38.

In Exercises 39–44, find the area of the triangle having theindicated angle and sides.

39.

40.

41.

42.

43.

44. b � 20a � 16,C � 84 30#,

c � 64a � 105,B � 72 30#,

c � 22b � 4.5,A � 5 15#,

c � 85b � 57,A � 43 45#,

c � 20a � 62,B � 130 ,

b � 6a � 4,C � 120 ,

a � 315.6A � 88 ,

a � 10.8A � 10 ,

a � 10A � 60 ,

a � 5A � 36 ,

b

b � 22a � 9.5,A � 25 4#,

a � b � 1A � 45 ,

b � 24a � 25,A � 120 ,

a � b � 25A � 120 ,

b � 12.8a � 4.5,A � 58 ,

b � 12.8a � 11.4,A � 58 ,

b � 21a � 34,A � 76 ,

b � 20a � 18,A � 76 ,

b � 200a � 125,A � 110 ,

b � 100a � 125,A � 110 ,

c � 50a � 35,C � 95.20 ,

b � 16a � 48,A � 110 15#,

c � 10a � 125,A � 100 ,

b � 4a � 14,A � 145 ,

c � 5.8b � 6.2,B � 2 45#,

b � 6.8a � 4.5,B � 15 30#,

c � 10a � 9,A � 60 ,

b � 5a � 8,A � 36 ,

a � 358C � 104 ,B � 28 ,

c �34B � 42 ,A � 55 ,

c � 16B � 45 ,A � 120 ,

c � 10B � 65 ,A � 35 ,

b � 4.8B � 8 15#,A � 5 40#,

c � 18.1C � 54.6 ,A � 83 20#,

c � 2.68C � 54.6 ,A � 24.3 ,

a � 21.6C � 16.7 ,A � 102.4 ,

A B

135°10°

b a

c = 45

C

A B

b a = 3.5

c25° 35°

C

A B40°35°

ab

c = 10

C

A B

ab = 20

c

45°

105°

C


45. HEIGHT Because of prevailing winds, a tree grew so

that it was leaning from the vertical. At a point

40 meters from the tree, the angle of elevation to the top

of the tree is (see figure). Find the height of the tree.

46. HEIGHT A flagpole at a right angle to the horizontal

is located on a slope that makes an angle of with the

horizontal. The flagpole’s shadow is 16 meters long and

points directly up the slope. The angle of elevation from

the tip of the shadow to the sun is

(a) Draw a triangle to represent the situation. Show the

known quantities on the triangle and use a variable

to indicate the height of the flagpole.

(b) Write an equation that can be used to find the height

of the flagpole.

(c) Find the height of the flagpole.

47. ANGLE OF ELEVATION A 10-meter utility pole casts

a 17-meter shadow directly down a slope when the

angle of elevation of the sun is (see figure). Find

the angle of elevation of the ground.

48. FLIGHT PATH A plane flies 500 kilometers with a

bearing of from Naples to Elgin (see figure). The

plane then flies 720 kilometers from Elgin to Canton

(Canton is due west of Naples). Find the bearing of the

flight from Elgin to Canton.

49. BRIDGE DESIGN A bridge is to be built across a

small lake from a gazebo to a dock (see figure). The

bearing from the gazebo to the dock is From a

tree 100 meters from the gazebo, the bearings to the

gazebo and the dock are and respec-

tively. Find the distance from the gazebo to the dock.

50. RAILROAD TRACK DESIGN The circular arc of a

railroad curve has a chord of length 3000 feet

corresponding to a central angle of

(a) Draw a diagram that visually represents the situation.

Show the known quantities on the diagram and use

the variables and to represent the radius of the

arc and the length of the arc, respectively.

(b) Find the radius of the circular arc.

(c) Find the length of the circular arc.

51. GLIDE PATH A pilot has just started on the glide path

for landing at an airport with a runway of length

9000 feet. The angles of depression from the plane to

the ends of the runway are and

(a) Draw a diagram that visually represents the

situation.

(b) Find the air distance the plane must travel until

touching down on the near end of the runway.

(c) Find the ground distance the plane must travel until

touching down.

(d) Find the altitude of the plane when the pilot begins

the descent.

52. LOCATING A FIRE The bearing from the Pine Knob

fire tower to the Colt Station fire tower is and

the two towers are 30 kilometers apart. A fire spotted by

rangers in each tower has a bearing of from

Pine Knob and from Colt Station (see figure).

Find the distance of the fire from each tower.

65°Fire

Pine Knob

Colt Station

80°70°

30 km

S

EW

N

Not drawn to scale

S 70 E

N 80 E

N 65 E,

18.8 .17.5

s

r

sr

40 .

74°Tree

Dock

Gazebo

100 m

28°

41°

S

EW

N

S 28 E,S 74 E

S 41 W.

44°

720 km 500 km

Naples

N

Canton

Elgin

S

EW

N

Not drawn to scale

316

42°

B

A

17 m

C

10 m

θ

θ

42° −

$,42

20 .

12

30°94°

h

40 m

h30

4

53. DISTANCE A boat is sailing due east parallel to the

shoreline at a speed of 10 miles per hour. At a given

time, the bearing to the lighthouse is S E, and

15 minutes later the bearing is S E (see figure). The

lighthouse is located at the shoreline. What is the

distance from the boat to the shoreline?

54. DISTANCE A family is traveling due west on a road

that passes a famous landmark. At a given time the bearing

to the landmark is N W, and after the family travels

5 miles farther the bearing is N W. What is the

closest the family will come to the landmark while on

the road?

55. ALTITUDE The angles of elevation to an airplane

from two points and on level ground are and

respectively. The points and are 2.2 miles apart,

and the airplane is east of both points in the same

vertical plane. Find the altitude of the plane.

56. DISTANCE The angles of elevation and to an

airplane from the airport control tower and from an

observation post 2 miles away are being continuously

monitored (see figure). Write an equation giving the

distance between the plane and observation post in

terms of and

EXPLORATION

TRUE OR FALSE? In Exercises 57–59, determine whetherthe statement is true or false. Justify your answer.

57. If a triangle contains an obtuse angle, then it must be

oblique.

58. Two angles and one side of a triangle do not necessarily

determine a unique triangle.

59. If three sides or three angles of an oblique triangle are

known, then the triangle can be solved.

60. GRAPHICAL AND NUMERICAL ANALYSIS In the

figure, and are positive angles.

(a) Write as a function of

(b) Use a graphing utility to graph the function in

part (a). Determine its domain and range.

(c) Use the result of part (a) to write as a function

of

(d) Use a graphing utility to graph the function in

part (c). Determine its domain and range.

(e) Complete the table. What can you infer?

FIGURE FOR 60 FIGURE FOR 61

61. GRAPHICAL ANALYSIS

(a) Write the area of the shaded region in the figure

as a function of

(b) Use a graphing utility to graph the function.

(c) Determine the domain of the function. Explain how

the area of the region and the domain of the function

would change if the eight-centimeter line segment

were decreased in length.

$.

A

30 cm

8 cm

20 cm

2

θ

θ

%.

c

%.&

%&

φA B

d

θ

Airport

control

towerObservation

post

2 mi Not drawn to scale

'.$

d

'$

BA72 ,

55 BA

38

62

70°63°

d S

EW

N

63

70


% 0.4 0.8 1.2 1.6 2.0 2.4 2.8

&

c

62. CAPSTONE In the figure, a triangle is to be formed

by drawing a line segment of length from to the

positive axis. For what value(s) of can you form

(a) one triangle, (b) two triangles, and (c) no triangles?

Explain your reasoning.

1 2 3 4 5

1

2

3

(0, 0)

(4, 3)

x

y

a

ax-

�4, 3�a

18 9

cα β

γ

Section 6.2 Law of Cosines 437

6.2 LAW OF COSINES


• Use the Law of Cosines to solveoblique triangles (SSS or SAS).

• Use the Law of Cosines to modeland solve real-life problems.

• Use Heron’s Area Formula to findthe area of a triangle.


You can use the Law of Cosines tosolve real-life problems involvingoblique triangles. For instance, inExercise 52 on page 443, you can usethe Law of Cosines to approximatehow far a baseball player has to run to make a catch.

Dan

iel Ben

djy/istock

photo.co

m

Introduction

Two cases remain in the list of conditions needed to solve an oblique triangle—SSS and

SAS. If you are given three sides (SSS), or two sides and their included angle (SAS),

none of the ratios in the Law of Sines would be complete. In such cases, you can use

the Law of Cosines.

For a proof of the Law of Cosines, see Proofs in Mathematics on page 488.

Three Sides of a Triangle—SSS

Find the three angles of the triangle in Figure 6.11.

FIGURE 6.11

Solution

It is a good idea first to find the angle opposite the longest side—side in this case.

Using the alternative form of the Law of Cosines, you find that

Because is negative, you know that is an obtuse angle given by

At this point, it is simpler to use the Law of Sines to determine

You know that must be acute because is obtuse, and a triangle can have, at most,

one obtuse angle. So, and

Now try Exercise 5.

41.12 .C 180 � 22.08 � 116.80 �A 22.08

BA

sin A � a�sin B

b 8�sin 116.80

19 0.37583

A.B 116.80 .

Bcos B

�0.45089.�82 � 142 � 192

2�8��14�cos B �

a2 � c2 � b2

2ac

b

AC

a = 8 ft c = 14 ft

b = 19 ft

B

Example 1

Law of Cosines

Standard Form Alternative Form

cos C �a2 � b2 � c2

2abc2 � a2 � b2 � 2ab cos C

cos B �a2 � c2 � b2

2acb2 � a2 � c2 � 2ac cos B

cos A �b2 � c2 � a2

2bca2 � b2 � c2 � 2bc cos A

Do you see why it was wise to find the largest angle first in Example 1? Knowing

the cosine of an angle, you can determine whether the angle is acute or obtuse. That is,

for Acute

for Obtuse

So, in Example 1, once you found that angle was obtuse, you knew that angles

and were both acute. If the largest angle is acute, the remaining two angles are acute

also.

Two Sides and the Included Angle—SAS

Find the remaining angles and side of the triangle in Figure 6.12.

FIGURE 6.12

Solution

Use the Law of Cosines to find the unknown side in the figure.

Because meters, you now know the ratio and you can use the

reciprocal form of the Law of Sines to solve for

There are two angles between and whose sine is 0.7034, and

For

For

Because side is the longest side of the triangle, must be the largest angle of the

triangle. So, and

Now try Exercise 7.

C 110.3 .B 44.7

Cc

C2 180 � 25 � 135.3 � 19.7 .

B2 135.3 ,

C1 180 � 25 � 44.7 � 110.3 .

B1 44.7 ,

B2 180 � 44.7 � 135.3 .

B1 44.7 180 0

0.7034

� 9� sin 25

5.4072

sin B � b�sin A

a

sin B

b�

sin A

a

B.

�sin A��aa 5.4072

a 5.4072

a2 29.2375

a2 � 92 � 122 � 2�9��12� cos 25

a2 � b2 � c2 � 2bc cos A

a

A B

a

c = 12 m

b = 9 m

25°

C

Example 2

C

AB

90 < $ < 180 .cos $ < 0

0 < $ < 90 cos $ > 0


When solving an oblique

triangle given three sides, you

use the alternative form of the

Law of Cosines to solve for an

angle. When solving an oblique

triangle given two sides and their

included angle, you use the

standard form of the Law of

Cosines to solve for an unknown.


Applications

An Application of the Law of Cosines

The pitcher’s mound on a women’s softball field is 43 feet from home plate and the

distance between the bases is 60 feet, as shown in Figure 6.13. (The pitcher’s mound is

not halfway between home plate and second base.) How far is the pitcher’s mound from

first base?

Solution

In triangle (line bisects the right angle at ), and

Using the Law of Cosines for this SAS case, you have

So, the approximate distance from the pitcher’s mound to first base is

feet.


An Application of the Law of Cosines

A ship travels 60 miles due east, then adjusts its course northward, as shown in

Figure 6.14. After traveling 80 miles in that direction, the ship is 139 miles from its

point of departure. Describe the bearing from point to point

FIGURE 6.14

Solution

You have and So, using the alternative form of the Law of

Cosines, you have

So, and thus the bearing measured from due north

from point to point is

or


N 76.15 E.166.15 � 90 � 76.15 ,

CB

B arccos��0.97094� 166.15 ,

�0.97094.

�802 � 602 � 1392

2�80��60�

cos B �a2 � c2 � b2

2ac

c � 60.b � 139,a � 80,

BA

c = 60 mi

C

a = 80 miS

EW

N

b = 139 mi

C.B

Example 4

h 1800.3 42.43

1800.3.� 432 � 602 � 2�43��60� cos 45

h2 � f 2 � p2 � 2fp cos H

p � 60.f � 43,HHPH � 45 HPF,

Example 3

H

PF

60 ft 60 ft

60 ft p = 60 ft

f = 43 ft

h

45°

FIGURE 6.13


Heron’s Area Formula

The Law of Cosines can be used to establish the following formula for the area of a

triangle. This formula is called Heron’s Area Formula after the Greek mathematician

Heron (c. 100 B.C.).

For a proof of Heron’s Area Formula, see Proofs in Mathematics on page 489.

Using Heron’s Area Formula

Find the area of a triangle having sides of lengths meters, meters, and

meters.

Solution

Because , Heron’s Area Formula yields


You have now studied three different formulas for the area of a triangle.

Standard Formula:

Oblique Triangle:

Heron’s Area Formula: Area � s�s � a��s � b��s � c�

Area �12bc sin A �

12ab sin C �

12ac sin B

Area �12bh

1131.89 square meters.

� 84�41��31��12�

Area � s�s � a��s � b��s � c�

s � �a � b � c��2 � 168�2 � 84

c � 72

b � 53a � 43

Example 5

Heron’s Area Formula

Given any triangle with sides of lengths and the area of the triangle is

Area

where s � �a � b � c��2.

� s�s � a��s � b��s � c�

c,b,a,

The Area of a Triangle Use the most appropriate formula to find the area of each triangle below. Show your work and give your reasons for choosing each formula.

a. b.

c. d.

4 ft

5 ft

3 ft

4 ft

2 ft

4 ft

2 ft 3 ft

4 ft

2 ft

50°


HISTORICAL NOTE

Heron of Alexandria (c. 100 B.C.)was a Greek geometer and

inventor. His works describe how to find the areas of

triangles, quadrilaterals, regularpolygons having 3 to 12 sides,

and circles as well as the surfaceareas and volumes of

three-dimensional objects.




1. If you are given three sides of a triangle, you would use the Law of ________ to find the three angles of the triangle.

2. If you are given two angles and any side of a triangle, you would use the Law of ________ to solve the triangle.

3. The standard form of the Law of Cosines for is ________ .

4. The Law of Cosines can be used to establish a formula for finding the area of a triangle called ________ ________

Formula.


cos B �a2 � c2 � b2

2ac

In Exercises 5–20, use the Law of Cosines to solve thetriangle. Round your answers to two decimal places.

5. 6.

7. 8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

In Exercises 21–26, complete the table by solving theparallelogram shown in the figure. (The lengths of thediagonals are given by and )

21. 5 8 45

22. 25 35 120

23. 10 14 20

24. 40 60 80

25. 15 25 20

26. 25 50 35

In Exercises 27–32, determine whether the Law of Sines orthe Law of Cosines is needed to solve the triangle. Then solvethe triangle.

27.

28.

29.

30.

31.

32.

In Exercises 33–40, use Heron’s Area Formula to find the areaof the triangle.

33.

34.

35.

36.

37.

38.

39.

40. a �35, b �

58, c �

38

a � 1, b �12, c �

34

c � 2.45b � 0.75,a � 3.05,

c � 15.05b � 8.46,a � 12.32,

c � 52b � 52,a � 75.4,

c � 9b � 10.2,a � 2.5,

c � 25b � 36,a � 33,

c � 17b � 12,a � 8,

a � 160, B � 12 , C � 7

A � 42 , B � 35 , c � 1.2

a � 11, b � 13, c � 7

A � 24 , a � 4, b � 18

a � 10, b � 12, C � 70

a � 8, c � 5, B � 40

��

��

��

��

��

� ��

'$dcba

φ

θ

a

c

b

d

d.c

b �34a �

38,C � 101 ,

b �79a �

49,C � 43 ,

b � 2.15a � 7.45,C � 15 15#,

c � 37a � 37,B � 125 40#,

c � 9.5a � 6.2,B � 75 20#,

c � 30a � 40,B � 10 35#,

c � 14b � 3,A � 48 ,

c � 7b � 6,A � 120 ,

c � 1.25b � 0.75,a � 1.42,

c � 52b � 52,a � 75.4,

c � 72b � 25,a � 55,

c � 21b � 15,a � 11,

BA

a = 9

c

b = 4.5105°

C

A B

b = 15

c = 30

a30°

C

A B

b = 3 a = 7

c = 8

C

A Bc = 16

b = 12 a = 10

C


41. NAVIGATION A boat race runs along a triangular

course marked by buoys and The race starts

with the boats headed west for 3700 meters. The other

two sides of the course lie to the north of the first side,

and their lengths are 1700 meters and 3000 meters.

Draw a figure that gives a visual representation of the

situation, and find the bearings for the last two legs of

the race.

42. NAVIGATION A plane flies 810 miles from Franklin

to Centerville with a bearing of Then it flies

648 miles from Centerville to Rosemount with a

bearing of Draw a figure that visually represents

the situation, and find the straight-line distance and

bearing from Franklin to Rosemount.

43. SURVEYING To approximate the length of a marsh, a

surveyor walks 250 meters from point to point , then

turns and walks 220 meters to point (see figure).

Approximate the length of the marsh.

44. SURVEYING A triangular parcel of land has 115 meters

of frontage, and the other boundaries have lengths of

76 meters and 92 meters. What angles does the frontage

make with the two other boundaries?

45. SURVEYING A triangular parcel of ground has sides

of lengths 725 feet, 650 feet, and 575 feet. Find the

measure of the largest angle.

46. STREETLIGHT DESIGN Determine the angle in the

design of the streetlight shown in the figure.

47. DISTANCE Two ships leave a port at 9 A.M. One

travels at a bearing of N W at 12 miles per hour, and

the other travels at a bearing of S W at 16 miles per

hour. Approximate how far apart they are at noon

that day.

48. LENGTH A 100-foot vertical tower is to be erected on

the side of a hill that makes a angle with the

horizontal (see figure). Find the length of each of the

two guy wires that will be anchored 75 feet uphill and

downhill from the base of the tower.

49. NAVIGATION On a map, Orlando is 178 millimeters

due south of Niagara Falls, Denver is 273 millimeters

from Orlando, and Denver is 235 millimeters from

Niagara Falls (see figure).

(a) Find the bearing of Denver from Orlando.

(b) Find the bearing of Denver from Niagara Falls.

50. NAVIGATION On a map, Minneapolis is 165 millimeters

due west of Albany, Phoenix is 216 millimeters from

Minneapolis, and Phoenix is 368 millimeters from

Albany (see figure).

(a) Find the bearing of Minneapolis from Phoenix.

(b) Find the bearing of Albany from Phoenix.

51. BASEBALL On a baseball diamond with 90-foot sides,

the pitcher’s mound is 60.5 feet from home plate. How

far is it from the pitcher’s mound to third base?

MinneapolisMinneapolisAlbanyAlbany

PhoenixPhoenix

216 mm216 mm

165 mm165 mm

368 mm368 mm

Denver

Niagara Falls

Orlando

273 mm

235 mm

178 mm

Denver

Niagara Falls

Orlando

273 mm

235 mm

178 mm

100 ft

75 ft75 ft

6°

6

67

53

4 21

3

2

θ

$

C A

75°

220 m 250 m

B

AC

C75

BA

32 .

75 .

C.B,A,


52. BASEBALL The baseball player in center field is playing

approximately 330 feet from the television camera that

is behind home plate. A batter hits a fly ball that goes to

the wall 420 feet from the camera (see figure). The

camera turns to follow the play. Approximately how

far does the center fielder have to run to make the catch?

53. AIRCRAFT TRACKING To determine the distance

between two aircraft, a tracking station continuously

determines the distance to each aircraft and the angle

between them (see figure). Determine the distance

between the planes when miles, and

miles.

54. AIRCRAFT TRACKING Use the figure for Exercise

53 to determine the distance between the planes when

miles, and miles.

55. TRUSSES is the midpoint of the line segment in

the truss rafter shown in the figure. What are the lengths

of the line segments and

56. ENGINE DESIGN An engine has a seven-inch

connecting rod fastened to a crank (see figure).

(a) Use the Law of Cosines to write an equation giving

the relationship between and

(b) Write as a function of (Select the sign that

yields positive values of )

(c) Use a graphing utility to graph the function in

part (b).

(d) Use the graph in part (c) to determine the maximum

distance the piston moves in one cycle.


57. PAPER MANUFACTURING In a process with contin-

uous paper, the paper passes across three rollers of radii

3 inches, 4 inches, and 6 inches (see figure). The centers

of the three-inch and six-inch rollers are inches apart,

and the length of the arc in contact with the paper on the

four-inch roller is inches. Complete the table.

58. AWNING DESIGN A retractable awning above a patio

door lowers at an angle of from the exterior wall at

a height of 10 feet above the ground (see figure). No

direct sunlight is to enter the door when the angle of

elevation of the sun is greater than What is the

length of the awning?

59. GEOMETRY The lengths of the sides of a triangular

parcel of land are approximately 200 feet, 500 feet, and

600 feet. Approximate the area of the parcel.

70°

50°

10 ft

Suns raysx

x

70 .

50

s

d

ds

4 in.

6 in.

θ

3 in.7 in.1.5 in.

x

θ

x.

$.x

$.x

PS

Q

8 8 8 8

10

R

RS?QS,PQ,

PRQ

c � 20b � 20A � 11 ,

a

A

B

C

bc

a

c � 20

b � 35A � 42 ,

a

A

330 ft

420 ft

8°

8

(inches)d 9 10 12 13 14 15 16

(degrees)$

(inches)s


60. GEOMETRY A parking lot has the shape of a parallelo-

gram (see figure). The lengths of two adjacent sides are

70 meters and 100 meters. The angle between the two

sides is What is the area of the parking lot?

61. GEOMETRY You want to buy a triangular lot measuring

510 yards by 840 yards by 1120 yards. The price of the

land is $2000 per acre. How much does the land cost?

(Hint:

62. GEOMETRY You want to buy a triangular lot measuring

1350 feet by 1860 feet by 2490 feet. The price of the

land is $2200 per acre. How much does the land cost?

(Hint:

EXPLORATION

TRUE OR FALSE? In Exercises 63 and 64, determinewhether the statement is true or false. Justify your answer.

63. In Heron’s Area Formula, is the average of the lengths

of the three sides of the triangle.

64. In addition to SSS and SAS, the Law of Cosines can be

used to solve triangles with SSA conditions.

65. WRITING A triangle has side lengths of 10 centimeters,

16 centimeters, and 5 centimeters. Can the Law of

Cosines be used to solve the triangle? Explain.

66. WRITING Given a triangle with meters,

and can the Law of Cosines be

used to solve the triangle? Explain.

67. CIRCUMSCRIBED AND INSCRIBED CIRCLES Let

and be the radii of the circumscribed and inscribed

circles of a triangle respectively (see figure), and let

(a) Prove that

(b) Prove that

CIRCUMSCRIBED AND INSCRIBED CIRCLES In Exercises68 and 69, use the results of Exercise 67.

68. Given a triangle with and find

the areas of (a) the triangle, (b) the circumscribed circle,

and (c) the inscribed circle.

69. Find the length of the largest circular running track that

can be built on a triangular piece of property with sides

of lengths 200 feet, 250 feet, and 325 feet.

70. THINK ABOUT IT What familiar formula do you

obtain when you use the third form of the Law of

Cosines and you let

What is the relationship between the Law of

Cosines and this formula?

71. THINK ABOUT IT In Example 2, suppose

After solving for which angle would you solve for

next, or Are there two possible solutions for that

angle? If so, how can you determine which angle is the

correct solution?

72. WRITING Describe how the Law of Cosines can be

used to solve the ambiguous case of the oblique triangle

where feet, feet, and Is

the result the same as when the Law of Sines is used to

solve the triangle? Describe the advantages and the

disadvantages of each method.

73. WRITING In Exercise 72, the Law of Cosines was

used to solve a triangle in the two-solution case of SSA.

Can the Law of Cosines be used to solve the no-solution

and single-solution cases of SSA? Explain.

75. PROOF Use the Law of Cosines to prove that

76. PROOF Use the Law of Cosines to prove that

1

2bc�1 � cos A� �

a � b � c

2"a � b � c

2.

1

2bc�1 � cos A� �

a � b � c

2"

�a � b � c

2.

A � 20 .b � 30a � 12ABC,

C?B

a,

A � 115 .

C � 90 ?

c2 � a2 � b2 � 2ab cos C,

c � 72,b � 55,a � 25,

r ��s � a��s � b��s � c�s

.

2R �a

sin A�

b

sin B�

c

sin C.

BC a

b c

R

r

A

s �a � b � c

2.

ABC,

r

R

C � 110 ,A � 87 ,

b � 47

s

1 acre � 43,560 square feet)

1 acre � 4840 square yards)

70°

70 m

100 m

70 .

74. CAPSTONE Determine whether the Law of Sines or

the Law of Cosines is needed to solve the triangle.

(a) and (b) and

(c) and (d) and

(e) and (f) and ca, b,Cb, c,

cA, B,Ab, c,

Ca, c,aA, C,

Section 6.3 Vectors in the Plane 445

6.3 VECTORS IN THE PLANE


• Represent vectors as directed linesegments.

• Write the component forms of vectors.

• Perform basic vector operations and represent them graphically.

• Write vectors as linear combinationsof unit vectors.

• Find the direction angles of vectors.

• Use vectors to model and solve real-life problems.


You can use vectors to model andsolve real-life problems involving magnitude and direction. For instance,in Exercise 102 on page 457, you canuse vectors to determine the truedirection of a commercial jet.

Bill Bac

hman

/Pho

to Res

earche

rs, Inc.

Introduction

Quantities such as force and velocity involve both magnitude and direction and cannot be

completely characterized by a single real number. To represent such a quantity, you can

use a directed line segment, as shown in Figure 6.15. The directed line segment has

initial point and terminal point Its magnitude (or length) is denoted by and

can be found using the Distance Formula.

FIGURE 6.15 FIGURE 6.16

Two directed line segments that have the same magnitude and direction are

equivalent. For example, the directed line segments in Figure 6.16 are all equivalent.

The set of all directed line segments that are equivalent to the directed line segment

is a vector v in the plane, written Vectors are denoted by lowercase, boldface

letters such as , , and .

Vector Representation by Directed Line Segments

Let be represented by the directed line segment from to and let

be represented by the directed line segment from to as shown in

Figure 6.17. Show that and are equivalent.

FIGURE 6.17

Solution

From the Distance Formula, it follows that and have the same magnitude.

Moreover, both line segments have the same direction because they are both directed

toward the upper right on lines having a slope of

Because and have the same magnitude and direction, and are equivalent.


vuRS\

PQ\

4 � 2

4 � 1�

2 � 0

3 � 0�

2

3.

RS\

� �4 � 1�2 � �4 � 2�2 � 13 PQ\

� �3 � 0�2 � �2 � 0�2 � 13

RS\

PQ\

x

(1, 2)

(4, 4)

2 3 41(0, 0)

1

2

3

4

R

S

(3, 2)

Q

v

u

P

5

y

vu

S�4, 4�,R�1, 2�vQ�3, 2�,P�0, 0�u

Example 1

wvu

v � PQ\

.

PQ\

Initial pointP

Q

PQ

Terminal point

PQ\

Q.P

PQ\


Component Form of a Vector

The directed line segment whose initial point is the origin is often the most convenient

representative of a set of equivalent directed line segments. This representative of the

vector v is in standard position.

A vector whose initial point is the origin can be uniquely represented by the

coordinates of its terminal point This is the component form of a vector v,

written as The coordinates and are the components of . If both the

initial point and the terminal point lie at the origin, is the zero vector and is denoted

by

Two vectors and are equal if and only if and

For instance, in Example 1, the vector from to

is and the vector from to is

� !3, 2".!4 � 1, 4 � 2"v � RS\

�

S�4, 4�R�1, 2�vu � PQ\

� !3 � 0, 2 � 0" � !3, 2",Q�3, 2�P�0, 0�uu2 � v2.

u1 � v1v � !v1, v2"u � !u1, u2"

0 � !0, 0".v

vv2v1v � !v1, v2".�v1, v2�.

�0, 0�

Component Form of a Vector

The component form of the vector with initial point and terminal point

is given by

The magnitude (or length) of is given by

If is a unit vector. Moreover, if and only if is the zero vector 0.v v � 0v v � 1,

v � �q1 � p1�2 � �q2 � p2�2 � v12 � v2

2.

v

PQ\

� !q1 � p1, q2 � p2" � !v1, v2" � v.

Q�q1, q2�P�p1, p2�

TECHNOLOGY

You can graph vectors with agraphing utility by graphingdirected line segments. Consultthe user’s guide for your graphingutility for specific instructions.

Finding the Component Form of a Vector

Find the component form and magnitude of the vector that has initial point

and terminal point ��1, 5�.�4, �7�v

Example 2

Algebraic Solution

Let

and

Then, the components of are

So, and the magnitude of is


� 169 � 13.

v � ��5�2 � 122

vv � !�5, 12"

v2 � q2 � p2 � 5 � ��7� � 12.

v1 � q1 � p1 � �1 � 4 � �5

v � !v1, v2"

Q��1, 5� � �q1, q2�.

P�4, �7� � �p1, p2�

Graphical Solution

Use centimeter graph paper to plot the points and Carefully

sketch the vector Use the sketch to find the components of Then

use a centimeter ruler to find the magnitude of

FIGURE 6.18

Figure 6.18 shows that the components of are and so

Figure 6.18 also shows that the magnitude of is v � 13.vv � !�5, 12".v2 � 12,v1 � �5v

1cm

23

45

67

89

10111213

v.

v � !v1, v2".v.

Q��1, 5�.P�4, �7�


Vector Operations

The two basic vector operations are scalar multiplication and vector addition. In

operations with vectors, numbers are usually referred to as scalars. In this text, scalars

will always be real numbers. Geometrically, the product of a vector and a scalar is

the vector that is times as long as . If is positive, has the same direction as ,

and if is negative, has the direction opposite that of , as shown in Figure 6.19.

To add two vectors and geometrically, first position them (without changing

their lengths or directions) so that the initial point of the second vector coincides with

the terminal point of the first vector The sum is the vector formed by joining

the initial point of the first vector with the terminal point of the second vector as

shown in Figure 6.20. This technique is called the parallelogram law for vector

addition because the vector often called the resultant of vector addition, is the

diagonal of a parallelogram having adjacent sides and

FIGURE 6.20

The negative of is

Negative

and the difference of and is

Add See Figure 6.21.

Difference

To represent geometrically, you can use directed line segments with the same

initial point. The difference is the vector from the terminal point of to the

terminal point of , which is equal to as shown in Figure 6.21.u � ��v�,u

vu � v

u � v

� !u1 � v1, u2 � v2".

��v�.u � v � u � ��v�

vu

� !�v1, �v2"

�v � ��1�v

v � !v1, v2"

v

u

x

u+

v

y

v

u

x

y

v.u

u � v,

v,u

u � vu.

v

vu

vkvk

vkvkvkkv

Definitions of Vector Addition and Scalar Multiplication

Let and be vectors and let be a scalar (a real number).

Then the sum of and is the vector

Sum

and the scalar multiple of times is the vector

Scalar multipleku � k!u1, u2" � !ku1, ku2".

uk

u � v � !u1 � v1, u2 � v2"

vu

kv � !v1, v2"u � !u1, u2"

x

−v u − v

u

vu + (−v)

y

FIGURE 6.21

u � v � u � ��v�

v v v3122

2v −v −

FIGURE 6.19

The component definitions of vector addition and scalar multiplication are

illustrated in Example 3. In this example, notice that each of the vector operations can

be interpreted geometrically.

Vector Operations

Let and and find each of the following vectors.

a. 2 b. c.

Solution

a. Because you have

A sketch of 2 is shown in Figure 6.22.

b. The difference of and is

A sketch of is shown in Figure 6.23. Note that the figure shows the vector

difference as the sum

c. The sum of and 2 is

A sketch of is shown in Figure 6.24.


v � 2w

� !4, 13".

� !�2 � 6, 5 � 8"

� !�2, 5" � !6, 8"

� !�2, 5" � !2�3�, 2�4�"

v � 2w � !�2, 5" � 2!3, 4"

wv

w � ��v�.w � v

w � v

� !5, �1".

� !3 � ��2�, 4 � 5"

w � v � !3, 4" � !�2, 5"

vw

v

� !�4, 10".

� !2��2�, 2�5�"

2v � 2!�2, 5"

v � !�2, 5",

v � 2ww � vv

w � !3, 4",v � !�2, 5"

Example 3


x

−2 2−4−6−8

( 4, 10)−

( 2, 5)−

2v

v

4

6

8

10

y

FIGURE 6.22

x

3 4 5

(3, 4)

(5, −1)

4

3

2

1

−1

w

w − v

−v

y

FIGURE 6.23

x

−2−4−6 2 4 6 8

(−2, 5)

(4, 13)

2w

v

v + 2w

8

10

12

14

y

FIGURE 6.24


Vector addition and scalar multiplication share many of the properties of ordinary

arithmetic.

Property 9 can be stated as follows: the magnitude of the vector is the absolute

value of times the magnitude of

Unit Vectors

In many applications of vectors, it is useful to find a unit vector that has the same

direction as a given nonzero vector To do this, you can divide by its magnitude to

obtain

Unit vector in direction of

Note that is a scalar multiple of The vector has a magnitude of 1 and the same

direction as The vector is called a unit vector in the direction of

Finding a Unit Vector

Find a unit vector in the direction of and verify that the result has a

magnitude of 1.

Solution

The unit vector in the direction of is

This vector has a magnitude of 1 because


� �2

29 2

� � 5

29 2

� 4

29�

25

29�29

29� 1.

� $ �2

29,

5

29%.

�1

29!�2, 5"

v

v �

!�2, 5"��2�2 � �5�2

v

v � !�2, 5"

Example 4

v.uv.

uv.u

vu � unit vector �v

v � � 1

v v.

vv.

v.c

cv

Properties of Vector Addition and Scalar Multiplication

Let , , and be vectors and let and be scalars. Then the following properties

are true.

1. 2.

3. 4.

5. 6.

7. 8.

9. cv � c v

0�u� � 01�u� � u,c�u � v� � cu � cv

�c � d�u � cu � duc�du� � �cd �u

u � ��u� � 0u � 0 � u

�u � v� � w � u � �v � w�u � v � v � u

dcwvu

HISTORICAL NOTE

William Rowan Hamilton(1805–1865), an Irish

mathematician, did some of the earliest work with vectors.

Hamilton spent many yearsdeveloping a system of

vector-like quantities calledquaternions. Although Hamiltonwas convinced of the benefits ofquaternions, the operations hedefined did not produce good

models for physical phenomena.It was not until the latter half ofthe nineteenth century that theScottish physicist James Maxwell

(1831–1879) restructuredHamilton’s quaternions in a formuseful for representing physical

quantities such as force, velocity, and acceleration.

The Granger Collection

The unit vectors and are called the standard unit vectors and are

denoted by

and

as shown in Figure 6.25. (Note that the lowercase letter is written in boldface to

distinguish it from the imaginary number ) These vectors can be used to

represent any vector as follows.

The scalars and are called the horizontal and vertical components of

respectively. The vector sum

is called a linear combination of the vectors and . Any vector in the plane can be

written as a linear combination of the standard unit vectors and

Writing a Linear Combination of Unit Vectors

Let be the vector with initial point and terminal point Write as a

linear combination of the standard unit vectors and

Solution

Begin by writing the component form of the vector

This result is shown graphically in Figure 6.26.


Vector Operations

Let and let Find

Solution

You could solve this problem by converting and to component form. This, however,

is not necessary. It is just as easy to perform the operations in unit vector form.


� �12i � 19j

� �6i � 16j � 6i � 3j

2u � 3v � 2��3i � 8j� � 3�2i � j�

vu

2u � 3v.v � 2i � j.u � �3i � 8j

Example 6

� �3i � 8j

� !�3, 8"

u � !�1 � 2, 3 � ��5�"

u.

j.i

u��1, 3�.�2, �5�u

Example 5

j.i

ji

v1i � v2 j

v,v2v1

� v1i � v2 j

� v1!1, 0" � v2!0, 1"

v � !v1, v2"

v � !v1, v2",i � �1.

i

j � !0, 1"i � !1, 0"

!0, 1"!1, 0"


x

j = ⟨0, 1⟩

i = ⟨1, 0⟩

1 2

1

2

y

FIGURE 6.25

x

−8 −6 −4 −2 2 4 6

−2

−4

−6

4

6

8

(−1, 3)

(2, −5)

u

y

FIGURE 6.26


Direction Angles

If is a unit vector such that is the angle (measured counterclockwise) from the

positive -axis to , the terminal point of lies on the unit circle and you have

as shown in Figure 6.27. The angle is the direction angle of the vector

Suppose that is a unit vector with direction angle If is any vector

that makes an angle with the positive -axis, it has the same direction as and you

can write

Because it follows that the direction angle

for is determined from

Quotient identity

Multiply numerator and denominator by

Simplify.

Finding Direction Angles of Vectors

Find the direction angle of each vector.

a.

b.

Solution

a. The direction angle is

So, as shown in Figure 6.28.

b. The direction angle is

Moreover, because lies in Quadrant IV, lies in Quadrant IV and its

reference angle is

So, it follows that as shown in Figure 6.29.


& 360! � 53.13! � 306.87!,

� arctan�� 4

3 & �53.13! � 53.13!.

v � 3i � 4j

tan �b

a�

�4

3.

� 45!,

tan �b

a�

3

3� 1.

v � 3i � 4j

u � 3i � 3j

Example 7

�b

a.

v .� v sin

v cos

tan �sin

cos

v

v � a i � bj � v �cos �i � v �sin �j,

� v �cos �i � v �sin �j.

v � v !cos , sin "

ux

v � a i � bj .u

u.

u � !x, y" � !cos , sin " � �cos �i � �sin �j

uux

u

xθ

u

−1

−1

1

1

θx = cos

θy = sin

( , )x y

y

1

FIGURE 6.27

u �

x

1 2 3

1

2

3(3, 3)

u

= 45°θ

y

FIGURE 6.28

x

−1 2 3 41

1

−2

−1

−3

−4 (3, −4)

v

306.87°

y

FIGURE 6.29

Applications of Vectors

Finding the Component Form of a Vector

Find the component form of the vector that represents the velocity of an airplane

descending at a speed of 150 miles per hour at an angle below the horizontal, as

shown in Figure 6.30.

Solution

The velocity vector has a magnitude of 150 and a direction angle of

You can check that has a magnitude of 150, as follows.


Using Vectors to Determine Weight

A force of 600 pounds is required to pull a boat and trailer up a ramp inclined at 15

from the horizontal. Find the combined weight of the boat and trailer.

Solution

Based on Figure 6.31, you can make the following observations.

By construction, triangles and are similar. Therefore, angle is So,

in triangle you have

Consequently, the combined weight is approximately 2318 pounds. (In Figure 6.31,

note that is parallel to the ramp.)


AC\

BA\

& 2318.

BA\

�600

sin 15!

sin 15! �600

BA\

sin 15! � AC

\

BA

\

ABC

15!.ABCABCBWD

AC\

� force required to move boat up ramp � 600 pounds

BC\

� force against ramp

BA\

� force of gravity � combined weight of boat and trailer

!

Example 9

� 22,501.41 & 150

& 19,869.72 � 2631.69

v & ��140.96�2 � ��51.30�2

v

� !�140.96, �51.30"

& �140.96i � 51.30j

& 150��0.9397�i � 150��0.3420�j

� 150�cos 200!�i � 150�sin 200!�j

v � v �cos �i � v �sin �j

� 200!.v

20!

Example 8


x

200°

y

150

−50−150

−50

−100

FIGURE 6.30

AC

B

D15°

15°

W

FIGURE 6.31


Using Vectors to Find Speed and Direction

An airplane is traveling at a speed of 500 miles per hour with a bearing of at a fixed

altitude with a negligible wind velocity as shown in Figure 6.32(a). When the airplane

reaches a certain point, it encounters a wind with a velocity of 70 miles per hour in the

direction as shown in Figure 6.32(b). What are the resultant speed and

direction of the airplane?

(a) (b)

FIGURE 6.32

Solution

Using Figure 6.32, the velocity of the airplane (alone) is

and the velocity of the wind is

So, the velocity of the airplane (in the wind) is

and the resultant speed of the airplane is

miles per hour.

Finally, if is the direction angle of the flight path, you have

which implies that

So, the true direction of the airplane is approximately


270! � �180! � 112.6!� � 337.4!.

� 112.6!.& 180! � 67.4! & 180! � arctan��2.4065�

& �2.4065

tan &482.5

�200.5

& 522.5

v & ��200.5�2 � �482.5�2

& !�200.5, 482.5"

� !�250 � 352, 2503 � 352"v � v1 � v2

� !352, 352".v2 � 70!cos 45!, sin 45!"

� !�250, 2503"v1 � 500!cos 120!, sin 120!"

xx

v1

v2

vWind

y

θ

x

120°

v1

y

N 45! E,

330!

Example 10

Recall from Section 4.8 that in

air navigation, bearings can be

measured in degrees clockwise

from north.




1. A ________ ________ ________ can be used to represent a quantity that involves both magnitude and direction.

2. The directed line segment has ________ point and ________ point

3. The ________ of the directed line segment is denoted by

4. The set of all directed line segments that are equivalent to a given directed line segment is a

________ in the plane.

5. In order to show that two vectors are equivalent, you must show that they have the same ________

and the same ________ .

6. The directed line segment whose initial point is the origin is said to be in ________ ________ .

7. A vector that has a magnitude of 1 is called a ________ ________ .

8. The two basic vector operations are scalar ________ and vector ________ .

9. The vector is called the ________ of vector addition.

10. The vector sum is called a ________ ________ of the vectors and and the scalars and

are called the ________ and ________ components of respectively.


v,

v2v1j,iv1i � v2 j

u � v

v

PQ\

PQ\

.PQ\

Q.PPQ\

In Exercises 11 and 12, show that u and v are equivalent.

11. 12.

In Exercises 13–24, find the component form and themagnitude of the vector v.

13. 14.

15. 16.

17. 18.

Initial Point Terminal Point

19.

20.

21.

22.

23.

24.

In Exercises 25–30, use the figure to sketch a graph of thespecified vector. To print an enlarged copy of the graph, go tothe website www.mathgraphs.com.

25. 26. 5

27. 28.

29. 30. v �12uu � v

u � 2vu � v

v�v

x

uv

y

�9, 40��3, 11��15, 12��1, 5��9, 3��1, 11��8, �9��1, 3��5, �17��2, 7��5, 1��3, �5�

x

(3, −1)(−4, −1)4−5

−2−3−4−5

v

23

1

4

y

x

(3, 3)

(3, −2)

2 51 4−2 −1

−2

−3

v2

3

1

4

y

x

(−1, 4)

(2, 2)

21 3−2 −1−3

v

2

3

5

1

y

x

(−1, −1)

(3, 5)

42−2−4

v2

6

4

y

x

( 4, 2)− −

1

−3

−2

−1v−2−4 −3

y

x

(1, 3)

1−1 2 3

1

2

3

4

v

y

(0, −5)

(0, 4)

(3, 3)

42−4−2

−4

vu

4

x

y

(−3, −4)

(4, 1)(0, 0)

(6, 5)

(2, 4)

−2−2 2 4 6

v

u4

2

6

x

y


In Exercises 31–38, find (a) u v, (b) u v, and (c) 2u 3v, Then sketch each resultant vector.

31. 32.

33. 34.

35.

36.

37. 38.

In Exercises 39– 48, find a unit vector in the direction of thegiven vector. Verify that the result has a magnitude of 1.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

In Exercises 49–52, find the vector v with the givenmagnitude and the same direction as u.

Magnitude Direction

49.

50.

51.

52.

In Exercises 53–56, the initial and terminal points of a vectorare given. Write a linear combination of the standard unitvectors i and j.

Initial Point Terminal Point

53.

54.

55.

56.

In Exercises 57–62, find the component form of v and sketchthe specified vector operations geometrically, where

and

57. 58.

59. 60.

61. 62.

In Exercises 63–66, find the magnitude and direction angleof the vector v.

63. 64.

65.

66.

In Exercises 67–74, find the component form of v given itsmagnitude and the angle it makes with the positive -axis.Sketch v.

Magnitude Angle

67.

68.

69.

70.

71.

72.

73. in the direction

74. in the direction

In Exercises 75–78, find the component form of the sum of uand v with direction angles and

Magnitude Angle

75.

76.

77.

78.

In Exercises 79 and 80, use the Law of Cosines to find theangle between the vectors. ( Assume )

79.

80.

RESULTANT FORCE In Exercises 81 and 82, find the anglebetween the forces given the magnitude of their resultant.(Hint: Write force 1 as a vector in the direction of the positive-axis and force 2 as a vector at an angle with the positive -axis.)

Force 1 Force 2 Resultant Force

81. 45 pounds 60 pounds 90 pounds

82. 3000 pounds 1000 pounds 3750 pounds

83. VELOCITY A gun with a muzzle velocity of 1200 feet

per second is fired at an angle of above the horizontal.

Find the vertical and horizontal components of the

velocity.

84. Detroit Tigers pitcher Joel Zumaya was recorded throwing

a pitch at a velocity of 104 miles per hour. If he threw

the pitch at an angle of below the horizontal, find

the vertical and horizontal components of the velocity.

(Source: Damon Lichtenwalner, Baseball Info Solutions)

35!

6!

x

x

w � 2i � jv � i � 2j,

w � 2i � 2jv � i � j,

0! " ! " 180!.!

� 110! v v � 30

� 30! u u � 50

� 180! v v � 50

� 45! u u � 20

� 90! v v � 4

� 60! u u � 4

� 90! v v � 5

� 0! u u � 5

v. u

i � 3jv v � 2

3i � 4jv v � 3

� 90! v � 43

� 45! v � 23

� 150! v � 34

� 150! v � 72

� 45! v � 1

� 0! v � 3

x

v � 8�cos 135!i � sin 135!j �v � 3�cos 60!i � sin 60!j �

v � �5i � 4jv � 6i � 6j

v � u � 2wv �12�3u � w�

v � �u � wv � u � 2w

v �34 wv �

32u

w " i # 2j.u " 2i $ j,

�2, 3��1, �5��0, 1��6, 4��3, 6��0, �2��3, �2��2, 1�

u � !3, 3" v � 8

u � !2, 5" v � 9

u � !�12, �5" v � 3

u � !�3, 4" v � 10

w � 7j � 3iw � i � 2j

w � �6iw � 4j

v � 6i � 2jv � i � j

v � !5, �12"v � !�2, 2"u � !0, �2"u � !3, 0"

v � 3iu � 2j,v � ju � 2i,

v � 3ju � �2i � j,

v � 2i � 3ju � i � j,

v � !2, 1"u � !0, 0",v � !0, 0"u � !�5, 3",v � !4, 0"u � !2, 3",v � !1, 3"u � !2, 1",

$

$#


85. RESULTANT FORCE Forces with magnitudes of

125 newtons and 300 newtons act on a hook (see

figure). The angle between the two forces is Find

the direction and magnitude of the resultant of these

forces.



2000 newtons and 900 newtons act on a machine part at

angles of and respectively, with the -axis

(see figure). Find the direction and magnitude of the

resultant of these forces.

87. RESULTANT FORCE Three forces with magnitudes of

75 pounds, 100 pounds, and 125 pounds act on an

object at angles of and respectively, with

the positive -axis. Find the direction and magnitude of

the resultant of these forces.

88. RESULTANT FORCE Three forces with magnitudes of

70 pounds, 40 pounds, and 60 pounds act on an object

at angles of and respectively, with the

positive -axis. Find the direction and magnitude of the

resultant of these forces.

89. A traffic light weighing 12 pounds is suspended by two

cables (see figure). Find the tension in each cable.

90. Repeat Exercise 89 if and

CABLE TENSION In Exercises 91 and 92, use the figure todetermine the tension in each cable supporting the load.

91. 92.

93. TOW LINE TENSION A loaded barge is being towed

by two tugboats, and the magnitude of the resultant is

6000 pounds directed along the axis of the barge (see

figure). Find the tension in the tow lines if they each

make an angle with the axis of the barge.

94. ROPE TENSION To carry a 100-pound cylindrical

weight, two people lift on the ends of short ropes that

are tied to an eyelet on the top center of the cylinder.

Each rope makes a angle with the vertical. Draw a

figure that gives a visual representation of the situation,

and find the tension in the ropes.

In Exercises 95–98, a force of pounds is required to pull anobject weighing pounds up a ramp inclined at degreesfrom the horizontal.

95. Find if pounds and


97. Find if pounds and pounds.


99. WORK A heavy object is pulled 30 feet across a

floor, using a force of 100 pounds. The force is exerted

at an angle of above the horizontal (see figure).

Find the work done. (Use the formula for work,

where is the component of the force in the

direction of motion and is the distance.)


100. ROPE TENSION A tetherball weighing 1 pound is

pulled outward from the pole by a horizontal force

until the rope makes a angle with the pole (see

figure). Determine the resulting tension in the rope and

the magnitude of u.

45!

u

u

1 lb

Tension

45°50°

30 ft

100 lb

D

FW � FD,

50!

� 26!.W � 5000F

W � 15,000F � 5000

� 14!.F � 600W

� 12!.W � 100F

W

F

20!

18°

18°

18!

A

C

B

10 in. 20 in.

24 in.

5000 lb

A

C

B50° 30°

2000 lb

2 � 35!. 1 � 40!

1 = 20°θ 2 = 20°θ

x

135!,45!,�30!,

x

120!,45!,30!,

x�45!,30!

30°

−45°

900 newtons

x

2000 newtons

x300 newtons

125 newtons

45°

y

45!.


112. CAPSTONE The initial and terminal points of

vector are and respectively.

(a) Write in component form.

(b) Write as the linear combination of the standard

unit vectors and

(c) Sketch with its initial point at the origin.

(d) Find the magnitude of v.

v

j.i

v

v

�9, 1�,�3, �4�v

101. NAVIGATION An airplane is flying in the direction

of with an airspeed of 875 kilometers per hour.

Because of the wind, its groundspeed and direction are

800 kilometers per hour and respectively (see

figure). Find the direction and speed of the wind.

102. NAVIGATION A commercial jet is flying from

Miami to Seattle. The jet’s velocity with respect to the

air is 580 miles per hour, and its bearing is The

wind, at the altitude of the plane, is blowing from the

southwest with a velocity of 60 miles per hour.

(a) Draw a figure that gives a visual representation of

the situation.

(b) Write the velocity of the wind as a vector in

component form.

(c) Write the velocity of the jet relative to the air in

component form.

(d) What is the speed of the jet with respect to the

ground?

(e) What is the true direction of the jet?

EXPLORATION

TRUE OR FALSE? In Exercises 103–110, use the figure todetermine whether the statement is true or false. Justify youranswer.

103. 104.

105. 106.

107. 108.

109. 110.

111. PROOF Prove that is a unit

vector for any value of

113. GRAPHICAL REASONING Consider two forces

(a) Find as a function of

(b) Use a graphing utility to graph the function in part

(a) for

(c) Use the graph in part (b) to determine the range of

the function. What is its maximum, and for what

value of does it occur? What is its minimum, and

for what value of does it occur?

(d) Explain why the magnitude of the resultant is

never 0.

114. TECHNOLOGY Write a program for your graphing

utility that graphs two vectors and their difference

given the vectors in component form.

In Exercises 115 and 116, use the program in Exercise 114 tofind the difference of the vectors shown in the figure.

115. 116.

117. WRITING In your own words, state the difference

between a scalar and a vector. Give examples of each.

118. WRITING Give geometric descriptions of the

operations of addition of vectors and multiplication of

a vector by a scalar.

119. WRITING Identify the quantity as a scalar or as a

vector. Explain your reasoning.

(a) The muzzle velocity of a bullet

(b) The price of a company’s stock

(c) The air temperature in a room

(d) The weight of an automobile

x

(80, 80)

(10, 60)

(−20, 70)

(−100, 0) 50

−50

125

y

x

2 4 6 8

2

4

6

8(1, 6)

(5, 2)

(4, 5)

(9, 4)

y

0 " < 2�.

. F1 � F2

F1 � !10, 0" and F2 � 5!cos , sin ".

.

�cos �i � �sin �j

t � w � b � au � v � �2�b � t�a � d � 0a � w � �2d

v � w � �sa � u � c

c � sa � �d

ac s

d w

b

u

t

v

332!.

x

140°148°

875 kilometers per hour

800 kilometers per hour

y

Win

d

S

EW

N

140!,

148!,

The Dot Product of Two Vectors

So far you have studied two vector operations—vector addition and multiplication by a

scalar—each of which yields another vector. In this section, you will study a third

vector operation, the dot product. This product yields a scalar, rather than a vector.

For proofs of the properties of the dot product, see Proofs in Mathematics on page 490.

Finding Dot Products

Find each dot product.

a. b. c.

Solution

a.

b.

c.

Now try Exercise 7.

In Example 1, be sure you see that the dot product of two vectors is a scalar (a real

number), not a vector. Moreover, notice that the dot product can be positive, zero, or

negative.

� 0 � 6 � �6

!0, 3" # !4, �2" � 0�4� � 3��2�

� 2 � 2 � 0

!2, �1" # !1, 2" � 2�1� � ��1��2�

� 23

� 8 � 15

!4, 5" # !2, 3" � 4�2� � 5�3�

!0, 3" # !4, �2"!2, �1" # !1, 2"!4, 5" # !2, 3"

Example 1


6.4 VECTORS AND DOT PRODUCTS


• Find the dot product of two vectorsand use the properties of the dotproduct.

• Find the angle between two vectorsand determine whether two vectorsare orthogonal.

• Write a vector as the sum of twovector components.

• Use vectors to find the work doneby a force.


You can use the dot product of twovectors to solve real-life problemsinvolving two vector quantities. Forinstance, in Exercise 76 on page 466,you can use the dot product to findthe force necessary to keep a sportutility vehicle from rolling down a hill.

Joe Raedle/Getty Images

Definition of the Dot Product

The dot product of and is

u # v � u1v1 � u2v2.

v � !v1, v2"u � !u1, u2"

Properties of the Dot Product

Let , , and be vectors in the plane or in space and let be a scalar.

1.

2.

3.

4.

5. c�u # v� � cu # v � u # cv

v # v � v 2

u # �v � w� � u # v � u # w

0 # v � 0

u # v � v # u

cwvu

Section 6.4 Vectors and Dot Products 459

Using Properties of Dot Products

Let and Find each dot product.

a.

b.

Solution

Begin by finding the dot product of and

a.

b.

Notice that the product in part (a) is a vector, whereas the product in part (b) is a scalar.

Can you see why?


Dot Product and Magnitude

The dot product of with itself is 5. What is the magnitude of ?

Solution

Because and it follows that


The Angle Between Two Vectors

The angle between two nonzero vectors is the angle between their

respective standard position vectors, as shown in Figure 6.33. This angle can be found

using the dot product.

For a proof of the angle between two vectors, see Proofs in Mathematics on page 490.

0 " " �, ,

�5.

u �u # u

u # u � 5, u 2 � u # u

uu

Example 3

� �28

� 2��14�

u # 2v � 2�u # v�

� !�14, 28"

�u # v�w � �14!1, �2"

� �14

� ��1��2� � 3��4�

u # v � !�1, 3" # !2, �4"

v.u

u # 2v

�u # v�w

w � !1, �2".v � !2, �4",u � !�1, 3",

Example 2

Angle Between Two Vectors

If is the angle between two nonzero vectors and then

cos �u # v

u v .

v,u

−

vu

Origin

θ

v u

FIGURE 6.33

Finding the Angle Between Two Vectors

Find the angle between and

Solution

The two vectors and are shown in Figure 6.34.

This implies that the angle between the two vectors is


Rewriting the expression for the angle between two vectors in the form

Alternative form of dot product

produces an alternative way to calculate the dot product. From this form, you can see

that because and are always positive, and will always have the same

sign. Figure 6.35 shows the five possible orientations of two vectors.

The terms orthogonal and perpendicular mean essentially the same thing—meeting

at right angles. Note that the zero vector is orthogonal to every vector u, because

0 u � 0.

cos !u v v u

u v � u v cos !

! � arccos27

534" 22.2#.

�27

534

�#4, 3$ #3, 5$

#4, 3$ #3, 5$

cos ! �u v

u v

!

v � #3, 5$.u � #4, 3$!

Example 4


u v

θ

cos 1

Opposite Direction

FIGURE 6.35

! � �

! � �

u

v

θ

1 cos 0

Obtuse Angle

<!<�

�

2< ! < �

u

v

θ

cos 0

90 Angle#

�!

! ��

2

u

v

θ

0 cos 1

Acute Angle

<!<

0 < ! <

�

2

u

v

0

cos 1

Same Direction

�!

! �

x

1 2 3 4 5 6

1

2

3

4

5

6

v = ⟨3, 5⟩

u = ⟨4, 3⟩

θ

y

FIGURE 6.34

Definition of Orthogonal Vectors

The vectors and are orthogonal if u v � 0.vu


Determining Orthogonal Vectors

Are the vectors and orthogonal?

Solution

Find the dot product of the two vectors.

Because the dot product is 0, the two vectors are orthogonal (see Figure 6.36).

FIGURE 6.36


Finding Vector Components

You have already seen applications in which two vectors are added to produce a

resultant vector. Many applications in physics and engineering pose the reverse

problem—decomposing a given vector into the sum of two vector components.

Consider a boat on an inclined ramp, as shown in Figure 6.37. The force due to

gravity pulls the boat down the ramp and against the ramp. These two orthogonal

forces, and are vector components of . That is,

Vector components of

The negative of component represents the force needed to keep the boat from rolling

down the ramp, whereas represents the force that the tires must withstand against the

ramp. A procedure for finding and is shown on the following page.

FIGURE 6.37

F

w2

w1

w2w1

w2

w1

FF � w1 � w2.

Fw2,w1

F

v = ⟨6, 4⟩

u = ⟨2, −3⟩

x

1 2 3 4 5 6 7

1

2

3

4

−1

−2

−3

y

� 0� 2�6� � ��3��4�u v � #2, �3$ #6, 4$

v � #6, 4$u � #2, �3$

Example 5

TECHNOLOGY

The graphing utility program,Finding the Angle Between TwoVectors, found on the website forthis text at academic.cengage.com,graphs two vectors and in standard position and finds the measure of the angle between them. Use theprogram to verify the solutions forExamples 4 and 5.

v #c, d$u #a, b$

From the definition of vector components, you can see that it is easy to find the

component once you have found the projection of onto . To find the projection,

you can use the dot product, as follows.

is a scalar multiple of

Take dot product of each side with

and are orthogonal.

So,

and

w1 � projvu � cv �u v

v 2v.

c �u v

v 2

vw2� c v 2 � 0

� cv v � w2 v

v.u v � �cv � w2� v

v.w1u � w1 � w2 � cv � w2

vuw2


Definition of Vector Components

Let and be nonzero vectors such that

where and are orthogonal and is parallel to (or a scalar multiple of) ,

as shown in Figure 6.38. The vectors and are called vector components of

. The vector is the projection of onto and is denoted by

The vector is given by w2 � u � w1.w2

w1 � projvu.

vuw1u

w2w1

vw1w2w1

u � w1 � w2

vu

w1

w2 u

vθ

is acute.

FIGURE 6.38

!

w2

w1

u

vθ

is obtuse.!

Projection of u onto v

Let and be nonzero vectors. The projection of onto is

projvu � �u v

v 2 v.

vuvu


Decomposing a Vector into Components

Find the projection of onto Then write as the sum of two

orthogonal vectors, one of which is

Solution

The projection of onto is

as shown in Figure 6.39. The other component, is

So,


Finding a Force

A 200-pound cart sits on a ramp inclined at as shown in Figure 6.40. What force is

required to keep the cart from rolling down the ramp?

Solution

Because the force due to gravity is vertical and downward, you can represent the

gravitational force by the vector

Force due to gravity

To find the force required to keep the cart from rolling down the ramp, project onto

a unit vector in the direction of the ramp, as follows.

Unit vector along ramp

Therefore, the projection of onto is

The magnitude of this force is 100, and so a force of 100 pounds is required to keep the

cart from rolling down the ramp.


� �100�3

2i �

1

2j .

� ��200��1

2 v� �F v�v

� �F v

v 2 vw1 � projvF

vF

v � �cos 30#�i � �sin 30#�j �3

2i �

1

2j

v

F

F � �200j.

30#,

Example 7

u � w1 � w2 � %6

5,

2

5& � %9

5, �

27

5 & � #3, �5$.

w2 � u � w1 � #3, �5$ � %6

5,

2

5& � %9

5,�

27

5 &.w2,

� � 8

40 #6, 2$ � %6

5,

2

5&w1 � projvu � �u v

v 2 vvu

projvu.

uv � #6, 2$.u � #3, �5$

Example 6

x

−1 1 2 3 4 5 6

1

2

−1

−2

−3

−4

−5

v = ⟨6, 2⟩

u = ⟨3, −5⟩

w1

w2

y

FIGURE 6.39

F

v w1

30°

FIGURE 6.40

Work

The work done by a constant force acting along the line of motion of an object is

given by

as shown in Figure 6.41. If the constant force is not directed along the line of motion,

as shown in Figure 6.42, the work done by the force is given by

Projection form for work

Alternative form of dot product

Force acts along the line of motion. Force acts at angle with the line of motion.

FIGURE 6.41 FIGURE 6.42

This notion of work is summarized in the following definition.

Finding Work

To close a sliding barn door, a person pulls on a rope with a constant force of 50 pounds

at a constant angle of as shown in Figure 6.43. Find the work done in moving the

barn door 12 feet to its closed position.

Solution

Using a projection, you can calculate the work as follows.

Projection form for work

foot-pounds

So, the work done is 300 foot-pounds. You can verify this result by finding the vectors

and and calculating their dot product.


PQ\

F

� 300�1

2�50��12�

� �cos 60#� F PQ\

W � proj PQ\F PQ

\

60#,

Example 8

!

P Q

F

Fθ

projPQ

P Q

F

� F PQ\

.

projPQ\F � �cos !� F � �cos !� F PQ

\

W � projPQ\F PQ

\

W

F

� F PQ\

W � �magnitude of force��distance�

FW


Definition of Work

The work done by a constant force as its point of application moves along

the vector is given by either of the following.

1. Projection form

2. Dot product formW � F PQ\

W � projPQ\F PQ

\

PQ\

FW

P Q

60°

F

12 ft

12 ft

projPQ

F

FIGURE 6.43




1. The ________ ________ of two vectors yields a scalar, rather than a vector.

2. The dot product of and is ________ .

3. If is the angle between two nonzero vectors and then

4. The vectors and are ________ if

5. The projection of onto is given by

6. The work done by a constant force as its point of application moves along the vector

is given by or


W � ________ .W � ________

PQ\

FW

projvu � ________ .vu

u v � 0.vu

cos ! � ________ .v,u!

u v �v � #v1, v2$u � #u1, u2$

In Exercises 7–14, find the dot product of u and v.

7. 8.

9. 10.

11. 12.

13. 14.

In Exercises 15–24, use the vectors u v

and w to find the indicated quantity. Statewhether the result is a vector or a scalar.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

In Exercises 25–30, use the dot product to find the magnitudeof u.

25. 26.

27. 28.

29. 30.

In Exercises 31– 40, find the angle between the vectors.

31. 32.

33. 34.

35. 36.

37. 38.

39.

40.

In Exercises 41– 44, graph the vectors and find the degreemeasure of the angle between the vectors.

41. 42.

43. 44.

In Exercises 45– 48, use vectors to find the interior angles ofthe triangle with the given vertices.

45. 46.

47. 48.

In Exercises 49 –52, find u v, where is the angle betweenu and v.

49.

50.

51.

52. u � 4, v � 12, ! ��

3

u � 9, v � 36, ! �3�

4

u � 100, v � 250, ! ��

6

u � 4, v � 10, ! �2�

3

!

�7, 9��1, 9�,��3, 5�,��3, 0�, �2, 2�, �0, 6��8, 2��1, 7�,��3,�4�,�1, 2�, �3, 4�, �2, 5�

v � 8i � 3jv � �8i � 8j

u � 2i � 3ju � 5i � 5j

v � �4i � 4jv � �7i � 5j

u � 6i � 3ju � 3i � 4j

!

v � cos��2 i � sin��2 ju � cos��4 i � sin��4 jv � cos�3�

4 i � sin�3�

4 ju � cos��3 i � sin��3 j

v � 4i � 3jv � �6i � 6j

u � 2i � 3ju � 5i � 5j

v � �8i � 4jv � 6i � 4j

u � �6i � 3ju � 2i � j

v � i � 2jv � �2j

u � 2i � 3ju � 3i � 4j

v � #4, 0$v � #0, �2$u � #3, 2$u � #1, 0$

!

u � �21iu � 6j

u � 12i � 16ju � 20i � 25j

u � #4, �6$u � #�8, 15$

�v u� � �w v��u v� � �u w�2 � u w � 1

�u 2v�w�3w v�u�v u�w�u v�v3u vu u

#3,"1$ #"4, 2$, #3, 3$,

v � �2i � jv � �2i � 3j

u � i � 2ju � 3i � 2j

v � 7i � 2jv � i � j

u � 3i � 4ju � 4i � 2j

v � #�1, �8$v � #2, �3$u � #�2, 5$u � #�4, 1$v � #�2, 3$v � #�3, 2$u � #6, 10$u � #7, 1$


In Exercises 53–58, determine whether u and v are orthogonal,parallel, or neither.

53. 54.

55. 56.

57. 58.

In Exercises 59–62, find the projection of u onto v. Thenwrite u as the sum of two orthogonal vectors, one of which is

59. 60.

61. 62.

In Exercises 63–66, use the graph to determine mentally theprojection of u onto v. (The coordinates of the terminalpoints of the vectors in standard position are given.) Use theformula for the projection of u onto v to verify your result.

63. 64.

65. 66.

In Exercises 67–70, find two vectors in opposite directionsthat are orthogonal to the vector u. (There are many correctanswers.)

67. 68.

69. 70.

WORK In Exercises 71 and 72, find the work done inmoving a particle from to if the magnitude and directionof the force are given by v.

71.

72.

73. REVENUE The vector gives the

numbers of units of two models of cellular phones

produced by a telecommunications company. The vector

gives the prices (in dollars) of the

two models of cellular phones, respectively.

(a) Find the dot product and interpret the result in

the context of the problem.

(b) Identify the vector operation used to increase the

prices by 5%.

74. REVENUE The vector gives the

numbers of hamburgers and hot dogs, respectively, sold

at a fast-food stand in one month. The vector

gives the prices (in dollars) of the food

items.

(a) Find the dot product and interpret the result in

the context of the problem.

(b) Identify the vector operation used to increase the

prices by 2.5%.

75. BRAKING LOAD A truck with a gross weight of

30,000 pounds is parked on a slope of (see figure).

Assume that the only force to overcome is the force of

gravity.

(a) Find the force required to keep the truck from

rolling down the hill in terms of the slope

(b) Use a graphing utility to complete the table.

(c) Find the force perpendicular to the hill when

76. BRAKING LOAD A sport utility vehicle with a gross

weight of 5400 pounds is parked on a slope of

Assume that the only force to overcome is the force of

gravity. Find the force required to keep the vehicle from

rolling down the hill. Find the force perpendicular to

the hill.

10#.

d � 5#.

d.

d°

Weight = 30,000 lb

d #

u v

v � #2.25, 1.75$

u � #3140, 2750$

u v

v � #79.99, 99.99$

u � #4600, 5250$

P�1, 3�, Q��3, 5�, v � �2i � 3j

P�0, 0�, Q�4, 7�, v � #1, 4$

QP

u � �52 i � 3ju �

12 i �

23 j

u � #�8, 3$u � #3, 5$

2−2−2

−4

4 6

2

4(6, 4)

(2, −3)u

v

x

y

2−2−2

4 6

6(6, 4)

(−2, 3)

u

v

x

y

(6, 4)

(−3, −2)

x

y

u

v

−4 2 4 6

2

4

6

(6, 4)

(3, 2)

x

y

1 2 3 4 5 6−1

1

2

3

4

5

6

u

v

v � #�4, �1$v � #2, 15$u � #�3, �2$u � #0, 3$v � #1, �2$v � #6, 1$u � #4, 2$u � #2, 2$

projvu.

v � #sin !, �cos !$v � �i � j

u � #cos !, sin !$u � 2i � 2j

v � �2i � 2jv � 5i � 6j

u � iu �14�3i � j�

v � #�1, 5$v � # 12, �

54$

u � #3, 15$u � #�12, 30$

d 0# 1# 2# 3# 4# 5#

Force

d 6# 7# 8# 9# 10#

Force


77. WORK Determine the work done by a person lifting a

245-newton bag of sugar 3 meters.

78. WORK Determine the work done by a crane lifting a

2400-pound car 5 feet.

79. WORK A force of 45 pounds exerted at an angle of

above the horizontal is required to slide a table

across a floor (see figure). The table is dragged 20 feet.

Determine the work done in sliding the table.

80. WORK A tractor pulls a log 800 meters, and the

tension in the cable connecting the tractor and log is

approximately 15,691 newtons. The direction of the

force is above the horizontal. Approximate the work

done in pulling the log.

81. WORK One of the events in a local strongman contest

is to pull a cement block 100 feet. One competitor pulls

the block by exerting a force of 250 pounds on a rope

attached to the block at an angle of with the

horizontal (see figure). Find the work done in pulling

the block.

82. WORK A toy wagon is pulled by exerting a force of

25 pounds on a handle that makes a angle with the

horizontal (see figure). Find the work done in pulling

the wagon 50 feet.

83. PROGRAMMING Given vectors and in compo-

nent form, write a program for your graphing utility in

which the output is (a) (b) and (c) the angle

between and

84. PROGRAMMING Use the program you wrote in

Exercise 83 to find the angle between the given vectors.

(a) and

(b) and

85. PROGRAMMING Given vectors and in compo-

nent form, write a program for your graphing utility

in which the output is the component form of the

projection of onto

86. PROGRAMMING Use the program you wrote in

Exercise 85 to find the projection of onto for the

given vectors.

(a) and

(b) and

EXPLORATION


87. The work done by a constant force acting along the

line of motion of an object is represented by a vector.

88. A sliding door moves along the line of vector If

a force is applied to the door along a vector that is

orthogonal to then no work is done.

89. PROOF Use vectors to prove that the diagonals of a

rhombus are perpendicular.

91. THINK ABOUT IT What can be said about the vectors

and under each condition?

(a) The projection of onto equals .

(b) The projection of onto equals .

92. PROOF Prove the following.

93. PROOF Prove that if is a unit vector and is the

angle between and then

94. PROOF Prove that if is a unit vector and is the

angle between and then

u � cos��2 � ! i � sin��2 � ! j.j,u

!u

u � cos ! i � sin ! j.i,u

!u

u � v 2 � u 2 � v 2 � 2u v

0vu

uvu

vu

PQ\

,

PQ\

.

FW

v � #�2, 1$u � #3, �2$v � #�1, 3$u � #5, 6$

vu

v.u

vu

v � #4, 1$u � #2, �6$v � #2, 5$u � #8, �4$

v.u

v , u ,

vu

20°

20#

30°

100 ftNot drawn to scale

30#

35#

45 lb

30°

20 ft

30#

90. CAPSTONE What is known about , the angle

between two nonzero vectors and , under each

condition (see figure)?

(a) (b) (c) u v < 0u v > 0u v � 0

vu

Origin

θ

vu

!


6.5 TRIGONOMETRIC FORM OF A COMPLEX NUMBER


• Plot complex numbers in the complex plane and find absolutevalues of complex numbers.

• Write the trigonometric forms ofcomplex numbers.

• Multiply and divide complex numbers written in trigonometricform.

• Use DeMoivre’s Theorem to findpowers of complex numbers.

• Find nth roots of complex numbers.


You can use the trigonometric form of a complex number to perform operations with complex numbers. Forinstance, in Exercises 99–106 on page477, you can use the trigonometricforms of complex numbers to helpyou solve polynomial equations.

The Complex Plane

Just as real numbers can be represented by points on the real number line, you can

represent a complex number

as the point in a coordinate plane (the complex plane). The horizontal axis is

called the real axis and the vertical axis is called the imaginary axis, as shown in

Figure 6.44.

FIGURE 6.44

The absolute value of the complex number is defined as the distance

between the origin and the point

If the complex number is a real number (that is, if ), then this definition

agrees with that given for the absolute value of a real number

Finding the Absolute Value of a Complex Number

Plot and find its absolute value.

Solution

The number is plotted in Figure 6.45. It has an absolute value of

Now try Exercise 9.

� 29.

z � ��2�2 � 52

z � �2 � 5i

Example 1

a � 0i � a2 � 02 � a.

b � 0a � bi

�a, b�.�0, 0�a � bi

Real

axis1

3

2

1

−1

−2

2 3−2 −1−3

(3, 1)or3 + i

−2 − i(−2, −1) or

Imaginary

axis

�a, b�

z � a � bi

Definition of the Absolute Value of a Complex Number

The absolute value of the complex number is

a � bi � a2 � b2.

z � a � bi

1

3

4

5

2 3−2 −1−3−4 4

29

Real

axis

(−2, 5)

Imaginary

axis

FIGURE 6.45

Section 6.5 Trigonometric Form of a Complex Number 469

Trigonometric Form of a Complex Number

In Section 2.4, you learned how to add, subtract, multiply, and divide complex numbers.

To work effectively with powers and roots of complex numbers, it is helpful to write

complex numbers in trigonometric form. In Figure 6.46, consider the nonzero complex

number By letting be the angle from the positive real axis (measured

counterclockwise) to the line segment connecting the origin and the point you

can write

and

where Consequently, you have

from which you can obtain the trigonometric form of a complex number.

The trigonometric form of a complex number is also called the polar form. Because

there are infinitely many choices for the trigonometric form of a complex number is

not unique. Normally, is restricted to the interval although on occasion

it is convenient to use

Writing a Complex Number in Trigonometric Form

Write the complex number in trigonometric form.

Solution

The absolute value of is

and the reference angle is given by

Because and because lies in Quadrant III, you

choose to be So, the trigonometric form is

See Figure 6.47.


� 4�cos4�

3� i sin

4�

3 .z � r�cos ! � i sin !�

! � � � ��3 � 4��3.!

z � �2 � 23 itan��3� � 3

tan !$ �b

a�

�23

�2� 3.

!$

r � �2 � 23 i � ��2�2 � ��23 �2� 16 � 4

z

z � �2 � 23 i

Example 2

! < 0.

0 % ! < 2�,!

!,

a � bi � �r cos !� � �r sin !�i

r � a2 � b2.

b � r sin !a � r cos !

�a, b�,!a � bi.

Trigonometric Form of a Complex Number

The trigonometric form of the complex number is

where and The number

is the modulus of and is called an argument of z.!z,

rtan ! � b�a.a � r cos !, b � r sin !, r � a2 � b2,

z � r�cos ! � i sin !�

z � a � bi

θ

rb

a

(a ,b)

Real

axis

Imaginary

axis

FIGURE 6.46

1

−3

−2

−4

−2−33

4

z = 4

πReal

axis

z = −2 − 2 3 i

Imaginary

axis

FIGURE 6.47


Writing a Complex Number in Standard Form

Write the complex number in standard form

Solution

Because and you can write


Multiplication and Division of Complex Numbers

The trigonometric form adapts nicely to multiplication and division of complex numbers.

Suppose you are given two complex numbers

and

The product of and is given by

Using the sum and difference formulas for cosine and sine, you can rewrite this

equation as

This establishes the first part of the following rule. The second part is left for you to

verify (see Exercise 109).

Note that this rule says that to multiply two complex numbers you multiply

moduli and add arguments, whereas to divide two complex numbers you divide moduli

and subtract arguments.

z1z2 � r1r2�cos�!1 � !2� � i sin�!1 � !2��.

� r1r2��cos !1 cos !2 � sin!1 sin!2� � i�sin !1 cos!2 � cos !1 sin!2��.

z1z2 � r1r2�cos !1 � i sin !1��cos !2 � i sin !2�

z2z1

z2 � r2�cos !2 � i sin !2�.z1 � r1�cos !1 � i sin !1�

� 2 � 6 i.

� 22�1

2�3

2i

z � 8�cos��

3 � i sin��

3 �

sin��

3 � �3

2,cos��

3 �1

2

z � 8�cos��

3 � i sin��

3 �a � bi.

Example 3

TECHNOLOGY

A graphing utility can be used to convert a complex number in trigonometric (or polar) form to standard form. For specifickeystrokes, see the user’s manualfor your graphing utility.

Product and Quotient of Two Complex Numbers

Let and be complex numbers.

Product

Quotientz2 � 0z1

z2

�r1

r2

�cos�!1 � !2� � i sin�!1 � !2��,

z1z2 � r1r2�cos�!1 � !2� � i sin�!1 � !2��

z2 � r2�cos !2 � i sin !2�z1 � r1�cos !1 � i sin !1�


Multiplying Complex Numbers

Find the product of the complex numbers.

Solution

You can check this result by first converting the complex numbers to the standard

forms and and then multiplying algebraically, as in

Section 2.4.


Dividing Complex Numbers

Find the quotient of the complex numbers.

Solution


� �32

2�

32

2i

� 3��2

2 � i��2

2 �� 3�cos 225 � i sin 225 �

Divide moduli andsubtract arguments.�

24

8�cos�300 � 75 � � i sin�300 � 75 ��

z1

z2

�24�cos 300 � i sin 300 �

8�cos 75 � i sin 75 �

z2 � 8�cos 75 � i sin 75 �z1 � 24�cos 300 � i sin 300 �

z1�z2

Example 5

� 16i

� �43 � 4i � 12i � 43

z1z2 � ��1 � 3 i��43 � 4i�

z2 � 43 � 4iz1 � �1 � 3 i

� 16i

� 16�0 � i�1��

� 16�cos�

2� i sin

�

2

� 16�cos5�

2� i sin

5�

2

Multiply moduli and add arguments.� 16�cos�2�

3�

11�

6 � i sin�2�

3�

11�

6 �

z1z2 � 2�cos2�

3� i sin

2�

3 ! 8�cos11�

6� i sin

11�

6

z2 � 8�cos11�

6� i sin

11�

6 z1 � 2�cos2�

3� i sin

2�

3 z1z2

Example 4

TECHNOLOGY

Some graphing utilities can multiply and divide complexnumbers in trigonometric form.If you have access to such agraphing utility, use it to find and in Examples 4 and 5.z1�z2

z1z2

and are coterminal.�

2

5�

2


Powers of Complex Numbers

The trigonometric form of a complex number is used to raise a complex number to a

power. To accomplish this, consider repeated use of the multiplication rule.

This pattern leads to DeMoivre’s Theorem, which is named after the French mathe-

matician Abraham DeMoivre (1667–1754).

Finding Powers of a Complex Number

Use DeMoivre’s Theorem to find

Solution

First convert the complex number to trigonometric form using

and

So, the trigonometric form is

Then, by DeMoivre’s Theorem, you have


� 4096.

� 4096�1 � 0�

� 4096�cos 8� � i sin 8��

� 212�cos12�2��

3� i sin

12�2��3 �

��1 � 3 i�12� �2�cos

2�

3� i sin

2�

3 �12

z � �1 � 3 i � 2�cos2�

3� i sin

2�

3 .

" � arctan3

�1�

2�

3.r � ��1�2 � �3�2

� 2

��1 � 3i�12.

Example 6

. . .

z5 � r5�cos 5" � i sin 5"�

z4 � r4�cos 4" � i sin 4"�

z3 � r2�cos 2" � i sin 2"�r �cos " � i sin "� � r3�cos 3" � i sin 3"�

z2 � r�cos " � i sin "�r�cos " � i sin "� � r2�cos 2" � i sin 2"�

z � r�cos " � i sin "�

DeMoivre’s Theorem

If is a complex number and is a positive integer, then

� rn �cos n" � i sin n"�.

zn � �r �cos " � i sin "��n

nz � r�cos " � i sin "�

HISTORICAL NOTE

Abraham DeMoivre (1667–1754)is remembered for his work in probability theory and

DeMoivre’s Theorem. His bookThe Doctrine of Chances

(published in 1718) includes thetheory of recurring series and the

theory of partial fractions.

The Granger Collection


Roots of Complex Numbers

Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial

equation of degree has solutions in the complex number system. So, the equation

has six solutions, and in this particular case you can find the six solutions by

factoring and using the Quadratic Formula.

Consequently, the solutions are

and

Each of these numbers is a sixth root of 1. In general, an nth root of a complex number

is defined as follows.

To find a formula for an th root of a complex number, let be an th root of

where

and

By DeMoivre’s Theorem and the fact that you have

Taking the absolute value of each side of this equation, it follows that

Substituting back into the previous equation and dividing by you get

So, it follows that

and

Because both sine and cosine have a period of these last two equations have

solutions if and only if the angles differ by a multiple of Consequently, there must

exist an integer such that

By substituting this value of into the trigonometric form of you get the result stated

on the following page.

u,#

# �" � 2�k

n.

n# � " � 2�k

k

2�.

2�,

sin n# � sin ".cos n# � cos "

cos n# � i sin n# � cos " � i sin ".

r,

sn � r.

sn�cos n# � i sin n#� � r�cos " � i sin "�.

un � z,

z � r�cos " � i sin "�.

u � s�cos # � i sin #�

z,nun

x �1 ± 3 i

2.x �

�1 ± 3 i

2,x � ±1,

� �x � 1��x2 � x � 1��x � 1��x2 � x � 1� � 0

x6 � 1 � �x3 � 1��x3 � 1�

x6 � 1

nn

Definition of an nth Root of a Complex Number

The complex number is an nth root of the complex number if

z � un � �a � bi�n.

zu � a � bi


When exceeds the roots begin to repeat. For instance, if the angle

is coterminal with which is also obtained when

The formula for the th roots of a complex number has a nice geometrical

interpretation, as shown in Figure 6.48. Note that because the th roots of all have the

same magnitude they all lie on a circle of radius with center at the origin.

Furthermore, because successive th roots have arguments that differ by the

roots are equally spaced around the circle.

You have already found the sixth roots of 1 by factoring and by using the Quadratic

Formula. Example 7 shows how you can solve the same problem with the formula for

th roots.

Finding the nth Roots of a Real Number

Find all sixth roots of 1.

Solution

First write 1 in the trigonometric form Then, by the th root

formula, with and the roots have the form

So, for 1, 2, 3, 4, and 5, the sixth roots are as follows. (See Figure 6.49.)

Increment by


cos 5�

3� i sin

5�

3�

1

2�3

2i

cos 4�

3� i sin

4�

3� �

1

2�3

2i

cos � � i sin � � �1

cos 2�

3� i sin

2�

3� �

1

2�3

2i

2�

n�

2�

6�

�

3cos

�

3� i sin

�

3�

1

2�3

2i

cos 0 � i sin 0 � 1

k � 0,

61�cos0 � 2�k

6� i sin

0 � 2�k

6 � cos�k

3� i sin

�k

3.

r � 1,n � 6

n1 � 1�cos 0 � i sin 0�.

Example 7

n

n

2��n,n

nrnr,

zn

zn

k � 0."�n,

" � 2�n

n�

"

n� 2�

k � n,n � 1,k

Finding nth Roots of a Complex Number

For a positive integer the complex number has exactly

distinct th roots given by

where 1, 2, . . . , n � 1.k � 0,

nr�cos" � 2�k

n� i sin

" � 2�k

n n

nz � r�cos " � i sin "�n,

2

2

n

nr

n

π

πReal

axis

Imaginary

axis

FIGURE 6.48

+

1−1

Real

axis−1 + 0i 1 + 0i

1

2++ i

3

2

1

2

− i1

2

3

2− i

1

2

3

2−

i3

2−

Imaginary

axis

FIGURE 6.49


In Figure 6.49, notice that the roots obtained in Example 7 all have a magnitude

of 1 and are equally spaced around the unit circle. Also notice that the complex roots

occur in conjugate pairs, as discussed in Section 2.5. The distinct th roots of 1 are

called the nth roots of unity.

Finding the nth Roots of a Complex Number

Find the three cube roots of

Solution

Because lies in Quadrant II, the trigonometric form of is

By the formula for th roots, the cube roots have the form

Finally, for you obtain the roots

See Figure 6.50.


0.3660 � 1.3660i.

68�cos135 � 360 �2�

3� i sin

135 � 360 �2�3 � 2�cos 285 � i sin 285 �

�1.3660 � 0.3660i

68�cos135 � 360 �1�

3� i sin

135 � 360 �1�3 �2�cos 165 � i sin 165 �

� 1 � i

68�cos135 � 360 �0�

3� i sin

135 � 360 �0�3 � 2�cos 45 � i sin 45 �

k � 0, 1, and 2,

68�cos135 � 360 k

3� i sin

135º � 360 k

3 .n

" � arctan� 2

�2 � 135 � 8 �cos 135 � i sin 135 �.

z � �2 � 2i

zz

z � �2 � 2i.

Example 8

nn

A Famous Mathematical Formula The famous formula

is called Euler’s Formula, after the Swiss mathematician Leonhard Euler(1707–1783). Although the interpretation of this formula is beyond the scope of this text, we decided to include it because it gives rise to one of the most wonderfulequations in mathematics.

This elegant equation relates the five most famous numbers in mathematics —0, 1,e, and i—in a single equation. Show how Euler’s Formula can be used to derive

this equation. ,

e i ! 1 " 0

ea�bi " ea�cos b ! i sin b�


1 + i−1.3660 + 0.3660i

0.3660 − 1.3660i

Real

axis

Imaginary

axis

1−2

−2

−1

1

2

FIGURE 6.50

Note in Example 8 that the

absolute value of is

and the angle is given by

tan " �b

a�

2

�2� �1.

"

� 8

� ��2�2 � 22

r � �2 � 2iz




1. The ________ ________ of a complex number is the distance between the origin and the point

2. The ________ ________ of a complex number is given by where is the ________

of and is the ________ of

3. ________ Theorem states that if is a complex number and is a positive integer, then

4. The complex number is an ________ ________ of the complex number if


z � un � �a � bi�n.zu � a � bi

zn � rn�cos n" � i sin n"�.nz � r �cos " � i sin "�

z."z

rz � r�cos " � i sin "�,z � a � bi

�a, b�.�0, 0�a � bi

In Exercises 5–10, plot the complex number and find itsabsolute value.

5. 6.

7. 8.

9. 10.

In Exercises 11–14, write the complex number in trigonometricform.

11. 12.

13. 14.

In Exercises 15–32, represent the complex numbergraphically, and find the trigonometric form of the number.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 2 26. 4

27. 28.

29. 30.

31. 32.

In Exercises 33–42, find the standard form of the complexnumber. Then represent the complex number graphically.

33. 34.

35.

36.

37. 38.

39. 40.

41.

42.

In Exercises 43–46, use a graphing utility to represent thecomplex number in standard form.

43. 44.

45. 46.

In Exercises 47–58, perform the operation and leave theresult in trigonometric form.

47.

48.

49.

50.

51.

52.

53. 54.

55. 56.

57. 58.6�cos 40 � i sin 40 �

7�cos 100 � i sin 100 �12�cos 92 � i sin 92 �

2�cos 122 � i sin 122 �

5�cos 4.3 � i sin 4.3�4�cos 2.1 � i sin 2.1�

cos � � i sin �

cos��3� � i sin��3�

cos 120 � i sin 120

2�cos 40 � i sin 40 �3�cos 50 � i sin 50 �9�cos 20 � i sin 20 �

�cos 5 � i sin 5 ��cos 20 � i sin 20 ��cos 80 � i sin 80 ��cos 330 � i sin 330 ��12�cos 100 � i sin 100 ��45�cos 300 � i sin 300 ��53�cos 120 � i sin 120 �� 23�cos 30 � i sin 30 ��34�cos

�

3� i sin

�

3 ��4�cos3�

4� i sin

3�

4 ��2�cos

�

4� i sin

�

4 ��6�cos�

12� i sin

�

12 �

9�cos 58º � i sin 58º�2�cos 155 � i sin 155 �

10�cos2�

5� i sin

2�

5 5�cos�

9� i sin

�

9

9.75�cos�280º 30% � � i sin�280º 30%��5�cos�198 45% � � i sin�198 45% ��

8�cos�

2� i sin

�

2 7�cos 0 � i sin 0�

6�cos5�

12� i sin

5�

12 9

4�cos3�

4� i sin

3�

4 8�cos 225 � i sin 225 �

48�cos��30 � � i sin��30 ��5�cos 135 � i sin 135 �2�cos 60 � i sin 60 �

�9 � 210 i�8 � 53 i

8 � 3i5 � 2i

�3 � i22 � i

3 � i�7 � 4i

12i�5i

52�3 � i��2�1 � 3 i�4 � 43 i1 � 3 i

5 � 5i1 � i

Realaxis−1−2

3

−3

z i= 1 + 3−

Imaginaryaxis

Realaxis

z = −3 − 3i

Imaginaryaxis

−2−3

−2

−3

s

Realaxis2

z = 2− 2

4

−2

−4

−4−6

Imaginaryaxiaxis

Realaxis

1

1

2

3

4

2−1−2

Imaginary

z = 3i

�8 � 3i4 � 6i

�7�7i

5 � 12i�6 � 8i


In Exercises 59–64, (a) write the trigonometric forms of thecomplex numbers, (b) perform the indicated operation usingthe trigonometric forms, and (c) perform the indicated operation using the standard forms, and check your resultwith that of part (b).

59. 60.

61. 62.

63. 64.

In Exercises 65 and 66, represent the powers and graphically. Describe the pattern.

65. 66.

In Exercises 67–82, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result instandard form.

67. 68.

69. 70.

71. 72.

73. 74.

75. 76.

77. 78.

79. 80.

81. 82.

In Exercises 83–98, (a) use the formula on page 474 to findthe indicated roots of the complex number, (b) representeach of the roots graphically, and (c) write each of the rootsin standard form.

83. Square roots of

84. Square roots of

85. Cube roots of

86. Fifth roots of

87. Cube roots of

88. Cube roots of

89. Square roots of 90. Fourth roots of

91. Fourth roots of 16 92. Fourth roots of

93. Fifth roots of 1 94. Cube roots of 1000

95. Cube roots of 96. Fourth roots of

97. Fifth roots of 98. Sixth roots of

In Exercises 99–106, use the formula on page 474 to find allthe solutions of the equation and represent the solutionsgraphically.

99. 100.

101. 102.

103. 104.

105. 106.

EXPLORATION


107. Geometrically, the th roots of any complex number

are all equally spaced around the unit circle centered at

the origin.

108. The product of two complex numbers is zero only when the

modulus of one (or both) of the complex numbers is zero.

109. Given two complex numbers

and show that

110. Show that is the complex

conjugate of

111. Use the trigonometric forms of and in Exercise 110

to find (a) and (b)

112. Show that the negative of is

113. Show that is a ninth root of

114. Show that is a fourth root of

115. THINK ABOUT IT Explain how you can use

DeMoivre’s Theorem to solve the polynomial

equation [Hint: Write as

16�cos � � i sin ��.]�16x4 � 16 � 0.

�2.2�1�4�1 � i��1.

12�1 � 3i�

�z � r �cos�" � �� i sin�" � ��.z � r�cos " � i sin "�

z�z, z � 0.zz

zz

z � r�cos " � i sin "�.z � r�cos��"� � i sin��"��

z1

z2

�r1

r2

�cos�"1 � "2� � i sin�"1 � "2��.

z2 � 0,z2 � r2�cos "2 � i sin "2�,z1 � r1�cos "1� i sin "1�

zn

x 4 � �1 � i� � 0x3 � �1 � i� � 0

x6 � 64i � 0x 4 � 16i � 0

x3 � 27 � 0x5 � 243 � 0

x3 � 1 � 0x 4 � i � 0

64i4�1 � i��4�125

i

625i�25i

�42��1 � i��125

2 �1 � 3 i�

32�cos5�

6� i sin

5�

6

8�cos2�

3� i sin

2�

3 16�cos 60 � i sin 60 �5�cos 120 � i sin 120 �

�2�cos�

8� i sin

�

8 �6

�3�cos 15 � i sin 15 ��4

�5 � 4i�3�3 � 2i�5�cos 0 � i sin 0�20�5�cos 3.2 � i sin 3.2��4�2�cos

�

2� i sin

�

2 �8

�cos�

4� i sin

�

4 12

�3�cos 60 � i sin 60 ��4�5�cos 20 � i sin 20 ��34�1 � 3 i�32�3 � i�10

�3 � 2i�8��1 � i�6�2 � 2i�6�1 � i�5

z �1

2�1 � 3 i�z �

2

2�1 � i�

z4z3,z2,z,

1 � 3 i

6 � 3i

3 � 4i

1 � 3 i

3i�1 � 2 i��2i�1 � i��3 � i��1 � i��2 � 2i��1 � i�

116. CAPSTONE Use the graph of the roots of a complexnumber.

(a) Write each of the roots in trigonometric form.

(b) Identify the complex number whose roots are given.Use a graphing utility to verify your results.

(i) (ii)

Realaxis

Imaginaryaxis

45°45°

45°45°

3 3

3 3

30°30°

−11

Realaxis

Imaginaryaxis

2 2

2


CHAPTER SUMMARY6

What Did You Learn? Explanation/Examples ReviewExercises

Section 6.3

Section 6.2

Section 6.1

Use the Law of Sines to solve

oblique triangles (AAS or ASA)

(p. 428).

Use the Law of Sines to solve

oblique triangles (SSA) (p. 430).

Find the areas of oblique triangles

(p. 432).

Use the Law of Sines to model and

solve real-life problems (p. 433).

Use the Law of Cosines to solve

oblique triangles (SSS or SAS)

(p. 437).

Use the Law of Cosines to model

and solve real-life problems

(p. 439).

Use Heron’s Area Formula to find

the area of a triangle (p. 440).

Represent vectors as directed line

segments (p. 445).

Write the component forms of

vectors (p. 446).

Law of Sines


is acute. is obtuse.

If two sides and one opposite angle are given, three

possible situations can occur: (1) no such triangle exists

(see Example 4), (2) one such triangle exists (see

Example 3), or (3) two distinct triangles may satisfy the

conditions. (see Example 5).

The area of any triangle is one-half the product of the

lengths of two sides times the sine of their included angle.

That is,

The Law of Sines can be used to approximate the total

distance of a boat race course. (See Example 7.)

Law of Cosines


The Law of Cosines can be used to find the distance

between a pitcher’s mound and first base on a women’s

softball field. (See Example 3.)

Heron’s Area Formula: Given any triangle with sides of

lengths and the area of the triangle is

where

The component form of the vector with initial point

and terminal point is given by

PQ\

� !q1 � p1, q2 � p2" � !v1, v2" � v.

Q�q1, q2�P�p1, p2�

Initial pointP

PQ Terminal pointQ

s � �a � b � c��2.Area � s�s � a��s � b��s � c�,c,b,a,

cos C �a2 � b2 � c2


cos B �a2 � c2 � b2


cos A �b2 � c2 � a2


Area �12bc sin A �

12ab sin C �

12ac sin B.

AA

A B

ab

c

h

C

A B

ab

c

h

C

a

sin A�

b

sin B�

c

sin C.

c,b,a,ABC

1–12

1–12

13–16

17–20

21–30

35–38

39– 42

43, 44

45–50

Chapter Summary 479

What Did You Learn? Explanation/Examples ReviewExercises

Section 6.5

Section 6.4

Section 6.3

Perform basic vector operations

and represent them graphically

(p. 447).

Write vectors as linear

combinations of unit vectors

(p. 449).

Find the direction angles of

vectors (p. 451).

Use vectors to model and solve

real-life problems (p. 452).

Find the dot product of two vectors

and use the properties of the dot

product (p. 458).

Find the angle between two

vectors and determine whether two

vectors are orthogonal (p. 459).

Write a vector as the sum of two

vector components (p. 461).

Use vectors to find the work done

by a force (p. 464).

Plot complex numbers in the

complex plane and find absolute

values of complex numbers (p. 468).

Write the trigonometric forms of

complex numbers (p. 469).

Multiply and divide complex

numbers written in trigonometric

form (p. 470).

Use DeMoivre’s Theorem to find

powers of complex numbers

(p. 472).

Find th roots of complex

numbers (p. 473).

n

Let and be vectors and let be a

scalar (a real number).

The scalars and are the horizontal and vertical

components of respectively. The vector sum is

the linear combination of the vectors and

If then the direction angle is

So,

Vectors can be used to find the resultant speed and

direction of an airplane. (See Example 10.)

The dot product of and is

If is the angle between two nonzero vectors and then

Vectors and are orthogonal if

Many applications in physics and engineering require the

decomposition of a given vector into the sum of two vector

components. (See Example 7.)

The work done by a constant force as its point

of application moves along the vector is given by

either of the following.

1. 2.

A complex number can be represented

as the point in the complex plane. The horizontal axis is

the real axis and the vertical axis is the imaginary axis. The

absolute value of is

The trigonometric form of the complex number

is where

and

Let and

DeMoivre’s Theorem: If is a

complex number and is a positive integer, then

The complex number is an th root of the

complex number if z � un � �a � bi�n.z

nu � a � bi

zn � �r�cos " � i sin "��n � rn�cos n" � i sin n"�.n

z � r�cos " � i sin "�

z2 � 0z1�z2 � �r1�r2��cos�"1 � "2� � i sin�"1 � "2��,z1z2 � r1r2�cos�"1 � "2� � i sin�"1 � "2��

z2 � r2�cos "2 � i sin "2�.z1 � r1�cos "1 � i sin "1�

tan " � b�a.r � a2 � b2,

b � r sin ",a � r cos ",z � r�cos " � i sin "�z � a � bi

a � bi � a2 � b2.z � a � bi

�a, b�z � a � bi

W � F ! PQ\

W � #projPQ\F# #PQ

\

#

PQ\

FW

u ! v � 0.vucos " �u ! v#u# #v#

.

v,u"

u ! v � u1v1 � u2v2.

v � !v1, v2"u � !u1, u2"

" � 45 .tan " � 2�2 � 1.

u � 2i � 2j,

j.i

v1i � v2 jv,

v2v1

v � !v1, v2" � v1!1, 0" � v2!0, 1" � v1i � v2 j

u � v � !u1 � v1, u2 � v2"�v � !�v1, �v2"ku � !ku1, ku2"u � v � !u1 � v1, u2 � v2"

kv � !v1, v2"u � !u1, u2" 51–62

63–68

69–74

75–78

79–90

91–98

99–102

103–106

107–112

113–118

119, 120

121–124

125–134

In Exercises 1–12, use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Roundyour answers to two decimal places.

1. 2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

In Exercises 13–16, find the area of the triangle having theindicated angle and sides.

13.

14.

15.

16.

17. HEIGHT From a certain distance, the angle of

elevation to the top of a building is At a point

50 meters closer to the building, the angle of elevation

is Approximate the height of the building.

18. GEOMETRY Find the length of the side of the

parallelogram.

19. HEIGHT A tree stands on a hillside of slope from

the horizontal. From a point 75 feet down the hill, the

angle of elevation to the top of the tree is (see

figure). Find the height of the tree.

FIGURE FOR 19

20. RIVER WIDTH A surveyor finds that a tree on the

opposite bank of a river flowing due east has a bearing

of N E from a certain point and a bearing of

N W from a point 400 feet downstream. Find the

width of the river.

In Exercises 21–30, use the Law of Cosines to solve thetriangle. Round your answers to two decimal places.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

In Exercises 31–34, determine whether the Law of Sines orthe Law of Cosines is needed to solve the triangle. Then solvethe triangle.

31.

32.

33.

34. A � 44 , B � 31 , c � 2.8

a � 13, b � 15, c � 24

a � 4, c � 5, B � 52

b � 9, c � 13, C � 64

A � 62 , b � 11.34, c � 19.52

C � 43 , a � 22.5, b � 31.4

B � 150 , a � 10, c � 20

B � 108 , a � 11, c � 11

a � 16.4, b � 8.8, c � 12.2

a � 2.5, b � 5.0, c � 4.5

a � 75, b � 50, c � 110

a � 6, b � 9, c � 14

A Bc

b = 4100°

a = 7

C

A Bc = 17

a = 8b = 14

C

6.2

15

22 30%

45°

28°

75 ft

45

28

140°

16

w

12

w

31 .

17 .

A � 11 , b � 22, c � 21

C � 119 , a � 18, b � 6

B � 80 , a � 4, c � 8

A � 33 , b � 7, c � 10

B � 25 , a � 6.2, b � 4

A � 75 , a � 51.2, b � 33.7

B � 150 , a � 10, b � 3

B � 150 , b � 30, c � 10

B � 64 , C � 36 , a � 367

A � 24 , C � 48 , b � 27.5

A � 95 , B � 45 , c � 104.8

A � 16 , B � 98 , c � 8.4

B � 10 , C � 20 , c � 33

B � 72 , C � 82 , b � 54

A C121°

a = 19

22°b

cB

A C

a = 8

38°

70°

b

c

B

6.1


REVIEW EXERCISES See www.CalcChat.com for worked-out solutions to odd-numbered exercises.6

Review Exercises 481

35. GEOMETRY The lengths of the diagonals of a

parallelogram are 10 feet and 16 feet. Find the lengths

of the sides of the parallelogram if the diagonals

intersect at an angle of

36. GEOMETRY The lengths of the diagonals of a

parallelogram are 30 meters and 40 meters. Find the

lengths of the sides of the parallelogram if the diagonals

intersect at an angle of

37. SURVEYING To approximate the length of a marsh, a

surveyor walks 425 meters from point to point

Then the surveyor turns and walks 300 meters to

point (see figure). Approximate the length of the

marsh.

38. NAVIGATION Two planes leave an airport at approx-

imately the same time. One is flying 425 miles per hour

at a bearing of and the other is flying 530 miles

per hour at a bearing of Draw a figure that gives a

visual representation of the situation and determine the

distance between the planes after they have flown for

2 hours.

In Exercises 39–42, use Heron’s Area Formula to find the areaof the triangle.

39.

40.

41.

42.

In Exercises 43 and 44, show that and areequivalent.

43. 44.

In Exercises 45–50, find the component form of the vector vsatisfying the conditions.

45. 46.

47. Initial point: terminal point:

48. Initial point: terminal point:

49.

50.

In Exercises 51–58, find (a) (b) (c) and (d)

51.

52.

53.

54.

55.

56.

57.

58.

In Exercises 59–62, find the component form of and sketch the specified vector operations geometrically, where

and

59. 60.

61. 62.

In Exercises 63–66, write vector as a linear combination ofthe standard unit vectors and

63. 64.

65. has initial point and terminal point

66. has initial point and terminal point

In Exercises 67 and 68, write the vector in the form

67. 68.

In Exercises 69–74, find the magnitude and the directionangle of the vector

69.

70.

71. 72. v � �4i � 7jv � 5i � 4j

v � 3�cos 150 i � sin 150 j�v � 7�cos 60 i � sin 60 j�

v.

v � 4i � jv � �10i � 10j

#v #$cos # i ! sin #j%.v

�5, �9�.��2, 7�u

�9, 8�.�3, 4�u

u � !�6, �8"u � !�1, 5"

j.i

u

w �12vw � 3v

w � 4u � 5vw � 2u � v

v " 10i ! 3j.u " 6i $ 5j

w

u � �6j, v � i � j

u � 4i, v � �i � 6j

u � �7i � 3j, v � 4i � j

u � 2i � j, v � 5i � 3j

u � !1, �8", v � !3, �2"u � !�5, 2", v � !4, 4"u � !4, 5", v � !0, �1"u � !�1, �3", v � !�3, 6"

3v ! 5u.4u,u $ v,u 1 v,

" � 225 #v # �12,

" � 120 #v # � 8,

�15, 9��1, 5�;�7, 3��0, 10�;

( )4

6

(0, 1)x

v

2 4 6

7

26,

y

2

−2

(−5, 4)

2

−2−4

4

6

(2, −1)

x

v

y

(1, 4)

42−4

vu

4

2

x

y

(−1, −4)

(−3, 2)

(3, −2)

( 2, 1)−

(0, 2)−

(6, 3)

(4, 6)

6−2−2

v

u

4

6

x

y

vu6.3

a �45, b �

34, c �

58

c � 3.7b � 15.8,a � 12.3,

c � 10b � 8,a � 15,

a � 3, b � 6, c � 8

67 .

355 ,

A

65°

300 m 425 m

B

C

ACC

65

B.A

34 .

28 .


73. 74.


85 pounds and 50 pounds act on a single point. The

angle between the forces is Describe the resultant

force.

76. ROPE TENSION A 180-pound weight is supported by

two ropes, as shown in the figure. Find the tension in

each rope.

77. NAVIGATION An airplane has an airspeed of

430 miles per hour at a bearing of The wind velocity

is 35 miles per hour in the direction of N E. Find the

resultant speed and direction of the airplane.

78. NAVIGATION An airplane has an airspeed of

724 kilometers per hour at a bearing of The wind

velocity is 32 kilometers per hour from the west. Find

the resultant speed and direction of the airplane.

In Exercises 79– 82, find the dot product of and

79. 80.

81. 82.

In Exercises 83–90, use the vectors andto find the indicated quantity. State whether the

result is a vector or a scalar.

83. 84.

85. 86.

87. 88.

89. 90.

In Exercises 91–94, find the angle between the vectors.

91.

92.

93.

94.

In Exercises 95–98, determine whether and areorthogonal, parallel, or neither.

95. 96.

97. 98.

In Exercises 99–102, find the projection of onto Thenwrite as the sum of two orthogonal vectors, one of which is

99.

100.

101.

102.

WORK In Exercises 103 and 104, find the work done inmoving a particle from to if the magnitude and directionof the force are given by

103.

104.

105. WORK Determine the work done (in foot-pounds)

by a crane lifting an 18,000-pound truck 48 inches.

106. WORK A mover exerts a horizontal force of

25 pounds on a crate as it is pushed up a ramp that is

12 feet long and inclined at an angle of above the

horizontal. Find the work done in pushing the crate.

In Exercises 107–112, plot the complex number andfind its absolute value.

107. 108.

109. 110.

111. 112.

In Exercises 113–118, write the complex number in trigonometric form.

113. 114.

115. 116.

117. 118.

In Exercises 119 and 120, (a) write the two complex numbersin trigonometric form, and (b) use the trigonometric forms tofind and where

119.

120. z2 � 2�3 � i�z1 � �3�1 � i�,z2 � �10iz1 � 23 � 2i,

z2 % 0.z1/z2,z1z2

�33 � 3i�5 � 12i

5 � 12i5 � 5i

�74i

�2 � 2 i2 � 2 i

�10 � 4i5 � 3i

�6i7i

6.5

20

P��2, �9�, Q��12, 8�, v � 3i � 6j

P�5, 3�, Q�8, 9�, v � !2, 7"

v.QP

v � !�5, 2"u � !�3, 5",v � !1, �1"u � !2, 7",v � !10, 0"u � !5, 6",v � !�8, �2"u � !�4, 3",

projvu.u

v.u

v � 3i � 6jv � i � 2j

u � �2i � ju � �i

v � !�2, 4"v � !8, 3"u � !1

4, �12"u � !�3, 8"

vu

v � !4, 33"u � !3, 3",v � !�2, 1"u � !22, �4",

v � cos 300 i � sin 300 j

u � cos 45 i � sin 45 j

v � cos5�

6i � sin

5�

6j

u � cos7�

4i � sin

7�

4j

#

�v ! v� � �v ! u��u ! u� � �u ! v��u ! v�vu�u ! v�#v#24 � #u#3u ! v2u ! u

v " !5, 1"u " !$4, 2"

v � 16i � 12jv � 11i � 5j

u � �7i � 2ju � 3i � 7j

v � !�4, �14"v � !�3, 9"u � !�7, 12"u � !6, 7"

v.u6.4

30 .

30

135 .

30° 30°

180 lb

15 .

v � 8i � jv � �3i � 3j

Review Exercises 483

In Exercises 121–124, use DeMoivre’s Theorem to find theindicated power of the complex number. Write the result instandard form.

121.

122.

123. 124.

In Exercises 125–128, (a) use the formula on page 474to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write eachof the roots in standard form.

125. Sixth roots of 126. Fourth roots of

127. Cube roots of 8 128. Fifth roots of

In Exercises 129–134, use the formula on page 474 to find all solutions of the equation and represent the solutionsgraphically.

129. 130.

131. 132.

133.

134.

EXPLORATION

TRUE OR FALSE? In Exercises 135–139, determinewhether the statement is true or false. Justify your answer.

135. The Law of Sines is true if one of the angles in the

triangle is a right angle.

136. When the Law of Sines is used, the solution is always

unique.

137. If is a unit vector in the direction of then

138. If then

139. is a solution of the equation

140. State the Law of Sines from memory.

141. State the Law of Cosines from memory.

142. What characterizes a vector in the plane?

143. Which vectors in the figure appear to be equivalent?

144. The vectors and have the same magnitudes in the

two figures. In which figure will the magnitude of the

sum be greater? Give a reason for your answer.

(a) (b)

145. Give a geometric description of the scalar multiple

of the vector for and for

146. Give a geometric description of the sum of the vectors

and

GRAPHICAL REASONING In Exercises 147 and 148, usethe graph of the roots of a complex number.

(a) Write each of the roots in trigonometric form.

(b) Identify the complex number whose roots are given. Usea graphing utility to verify your results.

147. 148.

149. The figure shows and Describe and

150. One of the fourth roots of a complex number is

shown in the figure.

(a) How many roots are not shown?

(b) Describe the other roots.

Realaxis

1−1

1z

30°

Imaginaryaxis

z

Realaxis

1−1

1

z1z2

θθ

Imaginaryaxis

z1�z2.z1z2z2.z1

60°

60°30°

30°4

4

44

3

3

Realaxis

Imaginaryaxis

60°

60°

4

4

4−2

−2

2

Realaxis

Imaginaryaxis

v.u

k < 0.k > 0u,

ku

x

uv

y

x

uv

y

vu

x

C BA

ED

y

x2 � 8i � 0.x � 3 � i

a � �b.v � a i � bj � 0,

v � v u.v,u

x5 � 4x3 � 8x2 � 32 � 0

x5 � x3 � x2 � 1 � 0

x4 � 64i � 0x3 � 8i � 0

x5 � 32 � 0x4 � 81 � 0

�1024

256i�729i

�1 � i�8�2 � 3i�6

�2�cos4�

15� i sin

4�

15 �5

�5�cos�

12� i sin

�

12 �4


CHAPTER TEST See www.CalcChat.com for worked-out solutions to odd-numbered exercises.6

Take this test as you would take a test in class. When you are finished, check your workagainst the answers given in the back of the book.

In Exercises 1–6, use the information to solve (if possible) the triangle. If two solutionsexist, find both solutions. Round your answers to two decimal places.

1. 2.

3. 4.

5. 6.

7. A triangular parcel of land has borders of lengths 60 meters, 70 meters, and

82 meters. Find the area of the parcel of land.

8. An airplane flies 370 miles from point to point with a bearing of It then

flies 240 miles from point to point with a bearing of (see figure). Find the

distance and bearing from point to point

In Exercises 9 and 10, find the component form of the vector satisfying the givenconditions.

9. Initial point of terminal point of

10. Magnitude of direction of

In Exercises 11–14, and Find the resultant vector and sketchits graph.

11. 12. 13. 14.

15. Find a unit vector in the direction of

16. Forces with magnitudes of 250 pounds and 130 pounds act on an object at angles

of and respectively, with the -axis. Find the direction and magnitude of

the resultant of these forces.

17. Find the angle between the vectors and

18. Are the vectors and orthogonal?

19. Find the projection of onto Then write as the sum of

two orthogonal vectors.

20. A 500-pound motorcycle is headed up a hill inclined at What force is required

to keep the motorcycle from rolling down the hill when stopped at a red light?

21. Write the complex number in trigonometric form.

22. Write the complex number in standard form.

In Exercises 23 and 24, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form.

23. 24.

25. Find the fourth roots of

26. Find all solutions of the equation and represent the solutions

graphically.

x3 � 27i � 0

256�1 � 3 i�.

�3 � 3i�6�3�cos7�

6� i sin

7�

6 �8

z � 6�cos 120 � i sin 120 �z � 5 � 5i

12 .

uv � !�5, �1".u � !6, 7"v � !5, 3"u � !6, �10"

v � !3, �2".u � !�1, 5"

x�60 ,45

u � !24, �7".

4u � 2v5u � 3vu � vu � v

v <!6, 5>.u <2, 7>

u � !3, �5"v: v � 12;v:

�11, �16�v:��3, 7�;v:

v

C.A

37 CB

24 .BA

C � 121 , a � 34, b � 55B � 100 , a � 15, b � 23

a � 4.0, b � 7.3, c � 12.4A � 24 , a � 11.2, b � 13.4

B � 110 , C � 28 , a � 15.6A � 24 , B � 68 , a � 12.2

A

B

24°

370 mi

37°

240 mi C

FIGURE FOR 8

Cumulative Test for Chapters 4–6 485

CUMULATIVE TEST FOR CHAPTERS 4–6 See www.CalcChat.com for worked-out solutions to odd-numbered exercises.6

Take this test as you would take a test in class. When you are finished, check your workagainst the answers given in the back of the book.

1. Consider the angle

(a) Sketch the angle in standard position.

(b) Determine a coterminal angle in the interval

(c) Convert the angle to radian measure.

(d) Find the reference angle

(e) Find the exact values of the six trigonometric functions of

2. Convert the angle radians to degrees. Round the answer to one decimal

place.

3. Find if and

In Exercises 4–6, sketch the graph of the function. (Include two full periods.)

4. 5. 6.

7. Find and such that the graph of the function matches

the graph in the figure.

8. Sketch the graph of the function over the interval

In Exercises 9 and 10, find the exact value of the expression without using a calculator.

9. 10.

11. Write an algebraic expression equivalent to

12. Use the fundamental identities to simplify:

13. Subtract and simplify:

In Exercises 14–16, verify the identity.

14.

15.

16.

In Exercises 17 and 18, find all solutions of the equation in the interval

17. 18.

19. Use the Quadratic Formula to solve the equation in the interval

20. Given that and angles and are both in Quadrant I, find

21. If find the exact value of

22. If find the exact value of sin !

2.tan ! �

4

3,

tan�2!�.tan ! �12,

tan�u � v�.vucos v �

35,sin u �

1213,

sin2 x � 2 sin x � 1 � 0.

�0, 2��:

3 tan ! � cot ! � 02 cos2 " � cos " � 0

[0, 2"#.

sin2 x cos2 x �18�1 � cos 4x�

sin�x � y� sin�x � y� � sin2 x � sin2 y

cot2 $�sec2 $ � 1� � 1

sin ! � 1

cos !�

cos !

sin ! � 1.

cos��2 � x csc x.

sin�arccos 2x�.

tan�arcsin35�tan�arctan 4.9�

�3� % x % 3�.f �x� �12x sin x

h�x� � a cos�bx � c�cb,a,

h�x� � �sec�x � ��g�x� �1

2 tan�x �

�

2 f �x� � 3 � 2 sin �x

sin ! < 0.tan ! � � 2120cos !

! � �1.45

!.

!&.

�0 , 360 �.

! � �120 .

x1 3

4

−3−4

y

FIGURE FOR 7


23. Write the product as a sum or difference.

24. Write as a product.

In Exercises 25–28, use the information to solve the triangle shown in the figure. Roundyour answers to two decimal places.

25. 26.

27. 28.

In Exercises 29 and 30, determine whether the Law of Sines or the Law of Cosines is neededto solve the triangle. Then solve the triangle.

29. 30.

31. Two sides of a triangle have lengths 7 inches and 12 inches. Their included angle

measures Find the area of the triangle.

32. Find the area of a triangle with sides of lengths 30 meters, 41 meters, and 45 meters.

33. Write the vector as a linear combination of the standard unit vectors

and

34. Find a unit vector in the direction of

35. Find for and

36. Find the projection of onto Then write as the sum of two

orthogonal vectors.

37. Write the complex number in trigonometric form.

38. Find the product of Write the

answer in standard form.

39. Find the three cube roots of 1.

40. Find all the solutions of the equation

41. A ceiling fan with 21-inch blades makes 63 revolutions per minute. Find the angular

speed of the fan in radians per minute. Find the linear speed of the tips of the blades

in inches per minute.

42. Find the area of the sector of a circle with a radius of 12 yards and a central angle

of

43. From a point 200 feet from a flagpole, the angles of elevation to the bottom and top

of the flag are and respectively. Approximate the height of the flag to

the nearest foot.

44. To determine the angle of elevation of a star in the sky, you get the star in your line

of vision with the backboard of a basketball hoop that is 5 feet higher than your

eyes (see figure). Your horizontal distance from the backboard is 12 feet. What is

the angle of elevation of the star?

45. Write a model for a particle in simple harmonic motion with a displacement of

4 inches and a period of 8 seconds.

46. An airplane’s velocity with respect to the air is 500 kilometers per hour, with a

bearing of The wind at the altitude of the plane has a velocity of 50 kilometers

per hour with a bearing of N E. What is the true direction of the plane, and what

is its speed relative to the ground?

47. A force of 85 pounds exerted at an angle of above the horizontal is required to

slide an object across a floor. The object is dragged 10 feet. Determine the work

done in sliding the object.

60

60

30 .

18 ,16 45&

105 .

x5 � 243 � 0.

�4�cos 30 � i sin 30 ��6�cos 120 � i sin 120 ��.�2 � 2i

uv � !1, 5".u � !8, �2"v � i � 2j.u � 3i � 4ju ' v

v � i � j.

j.

iu � !7, 8"

99 .

a � 1.2, b � 10, C � 80 A � 45 , B � 26 , c � 20

a � 4.7, b � 8.1, c � 10.3A � 30 , C � 90 , b � 10

A � 30 , b � 8, c � 10A � 30 , a � 9, b � 8

cos 9x � cos 7x

5 sin 3�

4' cos

7�

4

12 feet

5 feet

FIGURE FOR 44

A

b a

c B

C

FIGURE FOR 25–28

487

PROOFS IN MATHEMATICS

Proof

Let h be the altitude of either triangle found in the figure above. Then you have

or

or

Equating these two values of you have

or

Note that and because no angle of a triangle can have a measure

of or In a similar manner, construct an altitude from vertex to side

(extended in the obtuse triangle), as shown at the left. Then you have

or

or

Equating these two values of you have

or

By the Transitive Property of Equality you know that

So, the Law of Sines is established.

a

sin A�

b

sin B�

c

sin C.

a

sin A�

c

sin C.a sin C � c sin A

h,

h � a sin C.sin C �h

a

h � c sin Asin A �h

c

ACB180 .0

sin B � 0sin A � 0

a

sin A�

b

sin B.a sin B � b sin A

h,

h � a sin B.sin B �h

a

h � b sin Asin A �h

b

Law of Sines (p. 428)


a

c BA

b

C

A is obtuse.

a

c BA

b

C

A is acute.

a

sin A�

b

sin B�

c

sin C.

c,b,a,ABCLaw of Tangents

Besides the Law of Sines and

the Law of Cosines, there is also

a Law of Tangents, which was

developed by Francois Viete

(1540–1603). The Law of

Tangents follows from the Law

of Sines and the sum-to-product

formulas for sine and is defined

as follows.

The Law of Tangents can be

used to solve a triangle when

two sides and the included

angle are given (SAS). Before

calculators were invented, the

Law of Tangents was used to

solve the SAS case instead of

the Law of Cosines, because

computation with a table of

tangent values was easier.

a � b

a � b�

tan��A � B��2�tan��A � B��2�

a

c BA

b

C

A is obtuse.

a

c BA

b

C

A is acute.

488

Proof

To prove the first formula, consider the top triangle at the left, which has three acute

angles. Note that vertex has coordinates Furthermore, has coordinates

where and Because is the distance from vertex to vertex

it follows that

Distance Formula

Square each side.

Substitute for x and y.

Expand.

Factor out

To prove the second formula, consider the bottom triangle at the left, which also

has three acute angles. Note that vertex has coordinates Furthermore,

has coordinates where and Because is the

distance from vertex to vertex it follows that

Distance Formula

Square each side.

Substitute for x and y.

Expand.

Factor out

A similar argument is used to establish the third formula.

sin2 B � cos2 B � 1b2 � a2 � c2 � 2ac cos B.

a2.b2 � a2�sin2 B � cos2 B� � c2 � 2ac cos B

b2 � a2 cos2 B � 2ac cos B � c2 � a2 sin2 B

b2 � �a cos B � c�2 � �a sin B�2

b2 � �x � c�2 � �y � 0�2

b � �x � c�2 � �y � 0�2

A,C

by � a sin B.x � a cos B�x, y�,C�c, 0�.A

sin2 A � cos2 A � 1a2 � b2 � c2 � 2bc cos A.

b2.a2 � b2�sin2 A � cos2 A� � c2 � 2bc cos A

a2 � b2 cos2 A � 2bc cos A � c2 � b2 sin2 A

a2 � �b cos A � c�2 � �b sin A�2

a2 � �x � c�2 � �y � 0�2

a � �x � c�2 � � y � 0�2

B,

Cay � b sin A.x � b cos A

�x, y�,C�c, 0�.B

Law of Cosines (p. 437)


cos C �a2 � b2 � c2


cos B �a2 � c2 � b2


cos A �b2 � c2 � a2


x

C x y= ( , )

B c= ( , 0)

y

x

b

A

a

c

y

x

C x y= ( , )

y

x

a

B

b

c

y

A = (c, 0)

489

Proof

From Section 6.1, you know that

Square each side.

Pythagorean Identity

Factor.

Using the Law of Cosines, you can show that

and

Letting these two equations can be rewritten as

and

By substituting into the last formula for area, you can conclude that

Area � s�s � a��s � b��s � c�.

1

2bc�1 � cos A� � �s � b��s � c�.

1

2bc�1 � cos A� � s�s � a�

s � �a � b � c��2,

1

2bc�1 � cos A� �

a � b � c

2'a � b � c

2.

1

2bc�1 � cos A� �

a � b � c

2'

�a � b � c

2

��1

2bc�1 � cos A��1

2bc�1 � cos A��.

�1

4b2c2�1 � cos2 A�

Area �1

4b2c2 sin2 A

�Area�2 �1

4b2c2 sin2 A

Area �1

2bc sin A

Heron’s Area Formula (p. 440)

Given any triangle with sides of lengths and the area of the triangle is

where s �a � b � c

2.

Area � s�s � a��s � b��s � c�

c,b,a,

Formula for the area of

an oblique triangle

Take the square root of

each side.

490

Proof

Let and let c be a scalar.

1.

2.

3.

4.

5.

Proof

Consider the triangle determined by vectors u, v, and as shown in the figure. By

the Law of Cosines, you can write

cos ! �u ' v u v

.

v 2 � 2u ' v � u 2 � u 2 � v 2 � 2 u v cos !

v ' v � u ' v � v ' u � u ' u � u 2 � v 2 � 2 u v cos !

�v � u� ' v � �v � u� ' u � u 2 � v 2 � 2 u v cos !

�v � u� ' �v � u� � u 2 � v 2 � 2 u v cos !

v � u 2 � u 2 � v 2 � 2 u v cos !

v � u,

� cu ' v

� !cu1, cu2" ' !v1, v2"

� �cu1�v1 � �cu2�v2

� c�u1v1 � u2v2�

c�u ' v� � c�!u1, u2" ' !v1, v2"�

� v 2� �v12 � v2

2�2v ' v � v1

2 � v22

� u ' v � u ' w� �u1v1 � u2v2� � �u1w1 � u2w2�

� u1v1 � u1w1 � u2v2 � u2w2

� u1�v1 � w1� � u2�v2 � w2�

u ' �v � w� � u ' !v1 � w1, v2 � w2"

� 00 ' v � 0 ' v1 � 0 ' v2

� v ' u� v1u1 � v2u2u ' v � u1v1 � u2v2

0 � !0, 0",w � !w1, w2",v � !v1, v2",u � !u1, u2",

Properties of the Dot Product (p. 458)

Let and be vectors in the plane or in space and let be a scalar.

1. 2.

3. 4.

5. c�u ' v� � cu ' v � u ' cv

v ' v � v 2u ' �v � w� � u ' v � u ' w

0 ' v � 0u ' v � v ' u

cwv,u,

Angle Between Two Vectors (p. 459)

If is the angle between two nonzero vectors and then cos ! �u ' v u v

.v,u!

−

vu

Origin

θ

v u

491

1. In the figure, a beam of light is directed at the blue

mirror, reflected to the red mirror, and then reflected

back to the blue mirror. Find the distance that the light

travels from the red mirror back to the blue mirror.

2. A triathlete sets a course to swim S E from a point on

shore to a buoy mile away. After swimming 300 yards

through a strong current, the triathlete is off course at a

bearing of S E. Find the bearing and distance the

triathlete needs to swim to correct her course.

3. A hiking party is lost in a national park. Two ranger

stations have received an emergency SOS signal from

the party. Station B is 75 miles due east of station A. The

bearing from station A to the signal is S E and the

bearing from station B to the signal is S W.

(a) Draw a diagram that gives a visual representation of

the problem.

(b) Find the distance from each station to the SOS signal.

(c) A rescue party is in the park 20 miles from station A

at a bearing of S E. Find the distance and the

bearing the rescue party must travel to reach the lost

hiking party.

4. You are seeding a triangular courtyard. One side of the

courtyard is 52 feet long and another side is 46 feet long.

The angle opposite the 52-foot side is

(a) Draw a diagram that gives a visual representation of

the situation.

(b) How long is the third side of the courtyard?

(c) One bag of grass seed covers an area of 50 square

feet. How many bags of grass seed will you need to

cover the courtyard?

5. For each pair of vectors, find the following.

(i) (ii) (iii)

(iv) (v) (vi)

(a) (b)

(c) (d)

6. A skydiver is falling at a constant downward velocity of

120 miles per hour. In the figure, vector represents the

skydiver’s velocity. A steady breeze pushes the skydiver

to the east at 40 miles per hour. Vector represents the

wind velocity.

(a) Write the vectors and in component form.

(b) Let Use the figure to sketch To print

an enlarged copy of the graph, go to the website

www.mathgraphs.com.

(c) Find the magnitude of What information does the

magnitude give you about the skydiver’s fall?

(d) If there were no wind, the skydiver would fall in a

path perpendicular to the ground. At what angle to

the ground is the path of the skydiver when the

skydiver is affected by the 40-mile-per-hour wind

from due west?

(e) The skydiver is blown to the west at 30 miles per

hour. Draw a new figure that gives a visual represen-

tation of the problem and find the skydiver’s new

velocity.

s.

s.s � u � v.

vu

u

v

−20 20 40 60

20

40

60

80

100

120

140

EW

Down

Up

v

u

v � !5, 5"v � !2, 3"u � !2, �4"u � !1,

12"

v � !3, �3"v � !�1, 2"u � !0, 1"u � !1, �1"

u � v

u � v v

v u

u u � v v u

65 .

80

75

60

Buoy

300 yd

25°

35° 34

mi

S

EW

N

35

34

25

O T Q

P

25°

6 ft

4.7 ft

Blue mirror

Red mirror

θ

θ

α α

PT

This collection of thought-provoking and challenging exercises further explores andexpands upon concepts learned in this chapter.

PROBLEM SOLVING

492

7. Write the vector in terms of and given that the

terminal point of bisects the line segment (see figure).

8. Prove that if is orthogonal to and then is

orthogonal to

for any scalars and (see figure).

9. Two forces of the same magnitude and act at

angles and respectively. Use a diagram to com-

pare the work done by with the work done by in

moving along the vector if

(a)

(b) and

10. Four basic forces are in action during flight: weight, lift,

thrust, and drag. To fly through the air, an object must

overcome its own weight. To do this, it must create an

upward force called lift. To generate lift, a forward

motion called thrust is needed. The thrust must be great

enough to overcome air resistance, which is called drag.

For a commercial jet aircraft, a quick climb is important

to maximize efficiency because the performance of an

aircraft at high altitudes is enhanced. In addition, it is

necessary to clear obstacles such as buildings and

mountains and to reduce noise in residential areas. In

the diagram, the angle is called the climb angle. The

velocity of the plane can be represented by a vector

with a vertical component (called climb speed)

and a horizontal component where is the

speed of the plane.

When taking off, a pilot must decide how much of the

thrust to apply to each component. The more the thrust

is applied to the horizontal component, the faster the

airplane will gain speed. The more the thrust is applied

to the vertical component, the quicker the airplane will

climb.

FIGURE FOR 10

(a) Complete the table for an airplane that has a speedof miles per hour.

(b) Does an airplane’s speed equal the sum of the

vertical and horizontal components of its velocity?

If not, how could you find the speed of an airplane

whose velocity components were known?

(c) Use the result of part (b) to find the speed of an

airplane with the given velocity components.

(i) miles per hour

miles per hour

(ii) miles per hour

miles per hour v cos ! � 149.634

v sin ! � 10.463

v cos ! � 149.909

v sin ! � 5.235

v � 100

Thrust

Drag

Velocity

Weight

Lift

Climb angle

θ

θ

v v cos !,

v sin !v

!

!2 � 30 .!1 � 60

!1 � �!2

PQ

F2F1

!2,!1

F2F1

wv

u

dc

cv � dw

uw,vu

w

v

u

w

v,uw

! 0.5 1.0 1.5 2.0 2.5 3.0

v sin !

v cos !

Documents

Teacher ToolboxPRO 2 - in Trigonometry 6v2.toolboxpro.org/secure/teachers/1162/pc6_2e.pdf · 2012-06-11 · oblique triangles (AAS or ASA). • Use the Law of Sines to solve oblique