Material AICLE. 4º de ESO: Trigonometry (Solucionario)

3Material AICLE. 4º de ESO: Trigonometry (Solucionario)

TRIGONOMETRY

- Key -

4 Material AICLE. 4º de ESO: Trigonometry (Solucionario)

1.

2.

3.

trigonometricangle measurechordsine functiontriangulationtrigonometric series


4.

5.

SINECOSINE

TANGENTCOSECANT

SECANTCOTANGENT

Opposite / AdjacentOpposite / HypotenuseHypotenuse / OppositeAdjacent / Hypotenuse

Adjacent / OppositeHypotenuse / Adjacent

SINE RULECc

Bb

Aa

sinsinsin==

side a divided by the sine of angle A equals side b divided by the sine of angle B equals side c divided by the sine of angle C

COSINE RULE Abccba cos2222 −+=

side a squared equals side b squared plus side c squared minus twice b times c times the cosine of angle A


7.A. For the figure given on the left, the value of sin C is

Answer c / bThe sine of an angle is defined as Opposite Side / Hypotenuse. Now for angle C, the oppositeside is c and the hypotenuse is b. Hence the correct answer is c/b.

B. From the figure given on the right, the value of sin A + cos A is

Answer (a + c)/bWe know sin A = a/b, and cos A = c/b. Hence sin A + cos A = (a + c)/b.

C. From the figure given on the left, the value of cos C is

Answer a / bWe know that cos of any angle = Base/Hypotenuse. Now for the angle C, the base is a andhypotenuse is b. So cos C = a/b.

D. For the figure given on the right, which of the following relationships is true :

Answer cot A = c / aBy definition, cot A = 1 / tan A = c / a.

E. From the figure given on the left, the value of cos C + sin A is

Answer 2a/bThe value of cos C = a/b. Similarly the value of sin A = a/b. Hence cos C + sin A = 2a/b.

F. Which of the following relationships is true:

Answer sin A / cos A = tan AThe expression sin A / cos A = tan A is a useful one to remember in trigonometry.

G. tan A / sin A =

Answer sec Atan A = sin A / cos A. Therefore tan A / sin A = 1 / cos A = sec A.

H. (sin A / tan A) + cos A =

Answer 2 cos AWe know tan A = sin A / cos A. Therefore (sin A / tan A) + cos A = cos A + cos A = 2 cos A.

I. cot A tan A =

Answer 1cot A = 1 / tan A. Hence cot A tan A = 1.Alternatively cot A = cos A/sin A and tan A= sin A/cos A. So cot A tan A = (cos A/sin A) (sin A/cos A) = 1.

J. From the figure, the value of cosec A + cot A is:

Answer (b + c)/aWe know cosec A = b/a and cot A = c/a. Hence cosec A + cot A = (b + c)/a.


8.

Example 1 Example 2 Example 3

K. Which of the following relationships is true:

Answer cos A sec A = 1By definition, sec A = 1 / cos A. So cos A sec A = 1 is true.

L. From the figure, the value of sin2 A + cos2 A is

Answer 1This question is a bit tricky. We know sin A = a/b and cos A = c/b. So sin2 A + cos2 A = (a2 + c2)/ b2. By Pythagoras Theorem, a2 + c2 = b2 for a right-angled triangle. Hence sin2 A + cos2 A = 1,which is a famous identity.

M. From the figure, the value of cot C + cosec C is

Answer (a + b)/ccot C is Base/Opposite Side and cosec C is Hypotenuse/Opposite Side. From thesedefinitions, the values of cot C and cosec C are given by a/c and b/c respectively. Hencethe answer is (a + b)/c.

N. cosec A / sec A =

Answer cot ABy definition, cosec A = 1 / sin A and sec A = 1 / cos A. So cosec A / sec A = cos A / sin A =cot A.

O. For the figure given on the right, the value of cot A is

Answer tan CThe value of cot A is c/a. Similarly the value of tan C is c/a. Hence cot A = tan C.


9.

Example 4 Example 5

A.

(a) a = 2, A = 30°, B = 40°b = 2.571, c = 3.75

(b) b = 5, B = 45°, C = 60°a = 7.044, c = 6.124

(c) c = 3, A = 37°, B = 54°a = 1.806, b = 2.427

B.

(a) a = 3, b = 5, A = 32°Two possible triangles:B = 62°, C = 86°, c = 5.647and B = 118°, C = 30°, c = 2.833

(b) b = 2, c = 4, C = 63°B = 27°, A = 88°, a = 4.487

(c) c = 2, a = 1, B = 108°b = 2.457, A = 23.2°, C = 51.9°

C.

(a) a = 1, b = 2, C = 30°c = 1.239, A = 23.8°, B = 126.2°

(b) a = 3, c = 4, B = 50°b = 3.094, A = 48°, C = 82°

(c) b = 5, c = 10, A = 30°a = 6.197, B = 23.8°, C = 126.2°

D.

(a) a = 2, b = 3, c = 4A = 29.0°, B = 46.6°, C = 104.5°

(b) a = 1, b = 1, c = 1.5A = 41.4°, B = 41.4°, C = 97.2°


11.

15.

16.

a) The distance of the man from the tower is 20.21 mb) The length of the string used by the little boy is l = 2 h = 2 (15) = 30 mc) The height of the second tower is 46.19 md) The distance of the ship from the lighthouse is 35.49 me) The velocity of the plane is given by V = distance covered / time takenV= DE / 60 = 19.25 m/s

tan A = 1.23A = 50.9 °

tan B = 2.56B = 68.7 °

sin C = 0.78C = 51.3 °

sin D = 0.527D = 31.8 °

cos E = 0.352E = 69.4 °

cos F = 0.725F = 43.5 °

tan G = 0.786G = 38.2 °tan H = 1.275H = 51.9 °

sin I = 0.468I = 27.9 °

sin J = 0.867J = 60.1 °

sinA cosA tanA secA cosecA cotA0º 0 1 0 1 none none30º 0.5 0.8660 0.5773 1.1547 2 1.732045º 0.7071 0.7071 1 1.4142 1,4142 160º 0.8660 0.5 1.7320 2 1.1547 0.577390º 1 0 none none 1 0


17.

Score

Requirement 0 1 2 3 4

Story No Story or logical sequence of thoughts. One sentence.

No real story.Lacks imaginationand thought. No

real application ofa trigonometry

problem. Spellingand grammar

mistakes.

Story lacks one ofthe following:imagination,

completesentences,

complete thought,but still involves a

trigonometryproblem. Somegrammar and

spelling mistakes.

Imaginative storycomprised of

mostly completesentences that

involves atrigonometry

problem. Somegrammar mistakes

or a fewmisspellings.

Imaginative storycomprised of

completesentences that

involves atrigonometryproblem. No

grammar mistakesor misspellings.

Picture ordrawing

Nothing clearlydefinable, or

understandable

Picture lacks atleast two of the

following: illustratesthe story, clarity,appropriate size,color, appropriate

subject matter.The picture is not

very visuallyappealing.

Picture lacks atleast one of the

following: illustratesthe story, clarity,appropriate size,and appropriatesubject matter.

The picture is stillvisually appealing

and has color.

Creative picture ordrawing thatsomewhat

illustrates the storyand the

trigonometryproblem. Mainly

clear, colorful andvisually appealing.Appropriate size

and topic.

Creative picture ordrawing that

clearly illustratesthe story and the

trigonometryproblem. Clear,

colorful andvisually appealing.Appropriate size

and topic.

Diagram oftriangle

No serious attempt to make diagram.

Lacks labels, units and accuracy to story.

Not drawn with astraight edge orcomputer, and

missing ONE of thefollowing: units,labels. Correctlydrawn to match

story, picture andsolution.

Clearly drawnusing a straightedge

or computer.Missing TWO of the

following: units,labels. Correctlydrawn to match



or computer.Missing ONE of the

following: units,labels. Correctlydrawn to match



or computer.Units included,

labeled with rightangle. Correctlydrawn to match


Calculations

Not neatly typed or written and

missing TWO of thefollowing: units, formulas,

eachstep of the problem, correct

answer.

Not neatly typed orwritten and missing

ONE of thefollowing: units,formulas, each

step of theproblem, correct

answer.

Neatly typed orwritten. Missing

TWO of thefollowing: units,formulas, each


answer.

Neatly typed orwritten. Missing

One of thefollowing: units,formulas, each


answer

Neatly typed orwritten. Formulaslisted. Each stepclearly outlinedand included.

Units included inanswer. Correct

answer to problem

Presentation Put together on lined paper or notebook paper.

Rushed,incomplete, notprofessional.

Not quiteprofessional. Looksrushed or quickly

put together.Lacks colors, title or

does not fitstandard sizerequirements

Neat andProfessional buthas ONE of the

following: Does notuse 3 colors, title

not included, not astandard size.

Professional look(no tape showing,neatly constructed

or drawn)Story should betyped or neatly

printedAt least 3 Colors Title should be

obvious and neatSize limit: 8½ x 11

inches or a standard sheet of

construction paper.

Documents

Material AICLE. 4º de ESO: Trigonometry (Solucionario)