Module PMR

Trigonometry 169

Summary of the trigonometry

1. Tangent of an acute angle 2. Sine of an acute angle 3. Cosine of an acute angle 4. The values of Tangent, Sine and Cosine

C A B

a. Sin θ = Hypotenuse

Opposite

b. Cos θ = Hypotenuse

Adjacent

c. Tan θ = Adjacent

Opposite

A. Calculating the value of sine, cosine and tangent of an angle. Example :

1. In the diagram, AC = 26 cm and BC = 24 cm. Find the value of sinθ . C

24 cm 26 cm

B A

Opposite Side Hypotenus

Adjacent Side

θ

CHAPTER 11 :TRIGONOMETRY

θ

Module PMR

Trigonometry 170

Solution:

Sin C = AC

AB

= 26

10

= 13

5

Exercise : Calculate the value of sin θ for each of the following triangles. 1). R θ

6 cm P Q 10 cm

2). C B 5 cm 13 cm A

3). A 15 cm B C 8 cm

4). R 25 cm P Q 24 cm

AB2 = AC2 – BC2

= 262 - 242

= 676 – 576 = 100

AB = 100 = 10 cm

θ θ

θ

θ

Module PMR

Trigonometry 171

Example

1. In the diagram, AB = 15 cm and AC = 9 cm. Find the value of cos θ .

C B 9 cm

15 cm A Solution:

Cos θ = AB

BC

= 15

12

= 5

4

Exercise : Calculate the value of cos θ for each of the following diagram. P 1). R 17cm Q

2). R 12 cm Q 5 cm P

BC2 = AB2 – AC2 = 152 - 92 = 225 – 81 = 144

BC = 144 = 12 cm

8 cm

θ θ

θ

Module PMR

Trigonometry 172

3). K M 15 cm 8 cm L

4). A 12 cm B 15 cm C

Example

1. Find the value of tan θ in the diagram as shown. P

10 cm Q 6 cm R Solution :

Tan θ = PQ

QR

= 8

6

= 4

3

θ

θ

θ

PQ2 = PR2 – QR2

= 102 - 62

= 100 – 36 = 64

PQ = 64 = 8 cm

Module PMR

Trigonometry 173

Exercise : Calculate the value of tan θ for each of the following diagram. 1). P 25 cm 7 cm Q R

2). A B 13 cm 12 cm C

3). A 15 cm B 9 cm C

4). L 5 cm 13 cm M 4 cm N P

θ

θ

θ

θ

Module PMR

Trigonometry 174

B. Calculating the lengths of sides of a triangle from the tangent, sine and cosine of an angle. Example : Find the value of x in the following triangle. B x cm A C 15 cm

tan θ = 5

3

Solution :

tan θ = 5

3

5

3=AC

BC

5

3

15=x

155

3 ×=x

cmx 9=

θ

Module PMR

Trigonometry 175

Exercise. Find the value of x in each of the following triangles. 1). A x cm B 10 cm C

tan 5

2=θ

2). P 6 cm x cm Q R

tan 2

1=θ

3).

tan 24

7=θ

4).

tan 24

7=θ

θ

θ

θ

C

B A

x cm 48 cm

x cm

14cm

C

B A

θ

Module PMR

Trigonometry 176

5). K L x cm 12 cm M

sin 3

1=θ

6). R x cm S 6 cm T

sin 5

3=θ

7).

sin 5

3=θ

8).

sin 5

3=θ

θ θ

θ

x cm

15 cm

C B

A

θ x cm

9 cm C B

A

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Trigonometry 177

9). R x cm S 14 cm T

cos 7

5=θ

10). A x cm 16 cm B C

cos 8

5=θ

11).

cos 13

5=θ

12).

cos 13

5=θ

θθ

θ

26 cm

x cm C B

A

θ

x cm

52 cm

C

B A

Module PMR

Trigonometry 178

Common Errors

1. Find the value of cos x in the diagram as shown.

Error Correct step

Cos BC

CD=θ

8

7=

Cos CE

AC=θ

13

12=

2. Find the length of AC if sin 13

5=θ

Error Correct Steps

Sin 13

5=θ

13

5=AC

BC θtan=

AC

BC

13

510 =AC

Sin 13

5=θ

13

5=AB

BC

13

510 =AB

7 cm

6 cm

E

D

C 8 cm B 4 cm A

θ

θ

10 cm

B

C A

Module PMR

Trigonometry 179

5

1310×=AC

cmAC 26=

5

1310×=AB

cmAB 26= 222 1026 −=AC 576= 576=AC cmAC 24=

Extra Exercise 1. In diagram below, PQR is a straight line and T is the midpoint of straight line QTS. (a) Given that tan x 1=� , calculate the length of QTS. (b) State the value of cos y � .

P Q S

S

T

4 cm 15 cm x y

Module PMR

Trigonometry 180

2. In diagram below, S is the midpoint of straight line TSQ.

(a) Given that cos y5

4=� , calculate the length of TQ.

(b) Find sin x � . 3. In diagram below, C is the midpoint of straight line BD.

Given that sin x13

5=� , find the length of DC.

T S

R

Q

P

y

x

8 cm

16 cm

x

A

B C D

5 cm

Module PMR

Trigonometry 181

4. Diagram below shows two right angled triangles, PQR and RST. PRS and QRT are straight line.

Given that cos x17

15=� and sin y5

4=� .

(a) Find the value of tan x. (b) Calculate the length, in cm, of QRT.

P

R T

S

Q

x

y

17 cm

10 cm

Module PMR

Trigonometry 182

5. Diagram below shows two right angled triangles, ABC and ACD.

It is given that cos x5

4=� and tan y3

2=� .

(a) Find the value of sin x. (b) Calculate the length, in cm, of AD.

A B

C

D

x

y

8 cm

Module PMR

Trigonometry 183

6. Diagram below shows two right angled triangles, PQR and RST. PRS and QRT are straight lines.

It is given that sin x5

4=� and cos y13

12=� .

(a) Find the value of tan x. (b) Calculate the length, in cm, of QRT. 7. Diagram below show two right angled triangles, ABD and BCD.

Given that tan y5

12=� , find the value of cos x.

T

y

S R P

x

Q

13 cm

10 cm

D

A

B

C x

y

15 cm

13 cm

Module PMR

Trigonometry 184

8. In diagram below, JKL and KMN are straight line. M is the midpoint of the line KN.

Given that cos y5

4=� , find the value of sin x.

9. The diagram shows two right-angled triangles PQR and RSQ.

Given that sin x 13

5= ,

a). find the value of tan y, b). calculate the length of PR, in cm.

5 cm 8 cm

N

J

M

K L

y x

x

y 13 cm

16 cm Q P

R S

Module PMR

Trigonometry 185

10. In the diagram, tan x 5

12= .

Find the length of RS, in cm

x

12 cm 15 cm

T

S R Q

Module PMR

Trigonometry 186

PMR past year questions 2004 1). In Diagram 1, C is the midpoint of the straight line BD.

Find the value of tan xo

2005 2). Diagram 1 shows a right angled triangle EFG and DEF is a straight line. G 4 cm Find the value of cos x°

5 cm 13 cm

B C D

x

x°

2 cm D E F

Module PMR

Trigonometry 187

2006 3). Diagram 3 shows two right angled triangles, DAB and CDB.

It is given that tan y12

5=� and sin x2

1=� .

(a) Find the value of cos yo

(b) Calculate the length, in cm, of BC.

5 cm

B

C

y

x

Module PMR

Trigonometry 188

2007 4). Diagram 10 shows twi right angled triangles, PQT and SQR. PQR and TQS are straight lines. T

It is given that sin x13

5=� and cos y5

3=� .

(a) Find the value of tan x° (b) Calculate the length, in cm, of PQR.

15 cm

5 cm

y x P

Q R

S

Module PMR

Trigonometry 189

2008 5). Diagram 20 shows a right angled triangle PQR.

It is given that cos y13

5=� , calculate the length, in cm, of PQ.

y

P

Q R

39 cm

Module PMR

Trigonometry 190

CHAPTER 11 : TRIGONOMETRY ANSWERS

SIN θ

1). sin 5

4=θ

2). sin 13

12=θ

3). sin 17

8=θ

4). sin 25

7=θ

COS θ

1). cos 17

15=θ

2). cos 13

12=θ

3). cos 17

15=θ

4). cos 5

3=θ

TAN θ

1). tan 24

7=θ

2). tan 5

12=θ

3). tan 3

4=θ

4). tan 12

5=θ

VALUE OF x 1). x = 4 cm

2). x = 3 cm

3). x = 14 cm

4). x = 4.083 cm

5). x = 4 cm

6). x = 8 cm

7). x = 9 cm

8). x = 12 cm

9). x = 10 cm

10). x = 10 cm

11). x = 10 cm

12). x = 20 cm

EXTRA EXERCISE 1). (a) QTS = 8 cm

(b) Cos y17

15=�

2). (a) TQ = 20 cm

(b) Sin x5

3=�

3). DC = 6 cm

4). (a) Sin x17

8=�

(b) QRT = 14 cm

Module PMR

Trigonometry 191

5). (a) sin x5

3=�

(b) AD = 15 cm

6). (a) Tan x3

4=�

(b) QRT = 18 cm

7). Cos x5

3=�

8). Sin x13

12=�

9). (a) tan y 4

3=

(b) PR = 20 cm

10). RS = 4 cm

PMR past year questions 2004 2005

1). Tan x5

6=� 2). Cos x2

1−=�

2006 2007

3). a). Cos y13

12=� 4). a). tan x o = 12

5

b). BC = 26 cm b). PQR = 21 cm 2008 5). PQ = 15 cm