UNIT 4

UNIT 4 : TRIGONOMETRY

4.1 Trigonometric Ratios

LEARNING OUTCOMES The sides of a right–angled triangles are name the opposite side, adjacent
side and hypotenuse. The position of the opposite and adjacent sides
After completing the unit, students depends on the reference angle.
should be able to:
Hypotenuse B
1. Find the values of the o Opposite
trigonometric ratios of right- h
angled triangles
Aa
2. Find the values of trigonometric C
functions and inverse
trigonometric functions. Adjacent

3. Sketch the graph of y = sin  Figure 4.1
and y = cos 
Figure 4.1 shows a right-angled triangle where the reference angle is A, the
4. Compute unknown angles and adjacent side is b, the opposite side is a and the hypotenuse is c.
sides of oblique triangles by
using The trigonometric ratios are measurement of a right – angled triangle . The
• Sine rule three basic trigonometric ratios are sine , cosine and tangent. There are also
• Cosine rule other three common trigonometric ratios which are secant , cosecant and
cotangent.
5. Compute the area of oblique
triangles. The six trigonometric ratios below are related to Figure 4.1

Function Definition Function Definition

sin A opposite = o csc A hypotenuse = h
hypotenuse h opposite o

cos A adjacent = a sec A hypotenuse = h
hypotenuse h adjacent a

tan A opposite = o cot A adjacent = a
adjacent a opposite o

csc A = 1 sec A = 1 cot A = 1 Unit 4: Trigonometry
sin A cos A tan A

53

UNIT 4 : TRIGONOMETRY B
Example 4.1:

A

C

Figure 4.2

Figure 4.2 shows a right- angled triangle ABC . Name each sides of right-angled triangle as
opposite, adjacent or hypotenuse with the reference to the state angles.

a) Reference to ∡BAC
b) Reference to ∡ABC

Solution:

Angle Opposite Adjacent Hypotenuse
a) ∡ BAC BC AC AB
b) ∡ABC AC BC AB

Unit 4: Trigonometry

54

UNIT 4 : TRIGONOMETRY
Example 4.2:

P

8 cm
Q
6 cm

R
Figure 4.3

Figure 4.5 shows a right-angled triangles PQR with PQ = 8 cm and QR = 6 cm. Find value

of:

a) PR b) sin P c) tan R d) csc P e) sec R

Solution:

a) PR = √82 + 62 = √100 = 10 cm
b) sin P = 6 = 3

10 5
c) tan R = 8 = 4

63
d) csc P = 10 = 5

63
e) sec R = 10 = 5

63

Unit 4: Trigonometry

55

UNIT 4 : TRIGONOMETRY

Practice 4.1: 1) Figure 4.2 shows a right-angled triangle PQR.
P Name each side of right-angled triangle, reference to
the state angles.
Q
Angle Hypotenuse Opposite Adjacent
 RPQ QR
PRQ

R
Figure 4.2

2) Figure 4.3 shows a right-angled triangles ACB with B
AC = 4 cm and BC = 3 cm. Find the trigonometric 3 cm
ratios of the following: C

a) sin A = b) cos B =

A 4 cm
Figure 4.3
c) tan A= d) sec B=

A 3) Figure 4.4 shows a right-angled triangles ABC with
AC= 13 cm, AB = 5 cm and BC = 12 cm. Find the
5 cm 13 cm trigonometric ratios of the following:

B C a) =
12 cm
b) =
Figure 4.4 =

4) Figure 4.5 shows a right angle triangle ABC with A
B = 500. Determine the value for 7m

a) cos B = C 24 m B Unit 4: Trigonometry
b) sec B = Figure 4.5
56

UNIT 4 : TRIGONOMETRY 5) Evaluate the following trigonometric ratios in the Figure 4.6.
Q
a) tan P b) cot P

19 m

c) tan Q d) cot Q

P R
14 m

Figure 4.6

6) Find the values of trigonometric ratios in Figure 4.6. A
32.1 m
a) sin A b) cos C

B 19.2 m C

c) tan A d) csc C

Figure 4.7

4.2 Trigonometric Function

The expressions involving sin , cos , tan , csc , sec and cot are known as
trigonometric functions. A scientific calculator can be used to obtain the values of these
trigonometric functions easily.

Example 4.3

Use the calculator to evaluate the following:

a) sin 10o b) cos 143o c) tan 52 o Unit 4: Trigonometry

d) sec 222.10o e) cot 312.50

57

UNIT 4 : TRIGONOMETRY

Solution:

a) sin 100 = 0.1736 PRESS sin 10 = 0.1736
b) cos 1430 = -0.7986

c) tan 520 = 1.2799
d) sec 600 1

= =2 PRESS cos 60 = x-1 = 2
cos 60

e) cot 312.50 = tan 1 = - 0.9163
312.5

Match the equivalent trigonometric function with the value.

sec 450 0.866
tan 2000 1.414

csc 800 0.364
1.015
cos 300

4.2.2 Inverse Trigonometric Functions Unit 4: Trigonometry
If sin  = x , then  = sin−1 x
If cos  = x , then  = cos−1 x
If tan  = x , then  = tan −1 x

The expression sin−1 x , cos−1 x and tan −1 x is known as inverse trigonometric functions.
The values of the sin inverse trigonometric functions can also be easily obtained using
scientific calculators.

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UNIT 4 : TRIGONOMETRY

Example 4.4:

Evaluate the following:

a) sin−1 0.3241 b) cos−1 0.7582 c) tan−1 2.3120

Solution: PRESS SHIFT sin 0.3241 = 18. 910
a) sin-1 0.3241 = 18.910
b) cos-1 0.7582 = 40.690
c) tan-1 2.3120 = 66.61°

Find the value of the following angles, .
a.  = sin-1 0.9063
b.  = tan-1 5.6713
c.  = cos-1 -0.8660

Unit 4: Trigonometry

59

UNIT 4 : TRIGONOMETRY

Practice 4.2:
1. Find the value of the following trigonometric ratio.

a) sin 360 b) cos 530

c) tan 162.20 d) csc 3130

e) cot 1240 f) sec 2560

2. Find the value of the following angles, .

a)  = sin-1 0.866 b)  = cos-1 0.1736 c)  = tan-1 0.364

3. Determine the value of angles for each of following given functions:

a) cos  = 0.37604 b) tan  = 0.6945 c) sin  = 0.8304

4) Given tan = 6 , calculate the
8

x a) length of x b) angle

6 cm

8 cm c) sin d) cos Unit 4: Trigonometry
Figure 4.7 6
c
m

60

UNIT 4 : TRIGONOMETRY

3

5) If A is an angle of right angled triangle and cos A = , find

5

a) sin A b) tan A

6) in the diagram, R is the midpoint of the straight line
QRS. Find the value of

P S a) sin b) cos
R
12 cm
Q 10 cm

Unit 4: Trigonometry

61

UNIT 4 : TRIGONOMETRY y=

4.3 Graph of Trigonometric Function
Sketching and comparing graphs of sine and cosine. From Table 4.1, the graph of

sin θ is shown in Figure 4.7

 0 90 180 270 360
y = sin 0 1 0 -1 0

Table 4.1
y

1

x

90 180 270 360

−1 Figure 4.7

The shape of the graph y = sin x from x = 00 to x = 3600 is repeated for each complete
cycle. Hence the function y = sin x is periodic with the period of 3600.
The maximum and minimum values of the function y = sin x are 1 and -1 respectively.
These values are also called as amplitude.

From Table 4.2, the graph of y = cos θ is shown in Figure 4.8.

 0 90 180 270 360
y = cos  1 0 -1 0 1

y
Table 4.2

1

x

90 180 270 360

−1 Unit 4: Trigonometry

Figure 4.8 each
The shape of the graph of y = cos x from x = 00 to x = 3600 is repeated for
complete cycle. Hence the function y = cos x is periodic with a period of 360O.

The maximum and minimum value of the function y = cos x are 1 & -1 respectively.

62

UNIT 4 : TRIGONOMETRY
Practice 4.3 :

1. State the equation of the graph in each of the following figure: x
y x

a)
1

90 180 270 360

−1

y

1
b)

90 180 270 360

−1

2. Complete each of the following graphs for 0  x  360.
a) y
y = sin x
1

x

90 180 270 360

−1
b) y

1

x Unit 4: Trigonometry

90 180 270 360

−1
y = cos x

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UNIT 4 : TRIGONOMETRY

4.4 oblique Triangles

4.4.1 Sine Rule

B

c a a =b= c
sin A sinB sinC

AC
b

Using this rule, we must form a proportion by choosing two of the three ratios in which three

of the four terms are known.

This rule can be used when the following values are given :

a. Two angles and one side

b or a or

A B A BA c B

Two angles and Two angles and
non- included side included side

b. Two sides and one non – included angle

ba or ba

BA

Unit 4: Trigonometry

64

UNIT 4 : TRIGONOMETRY
Example 4.5:

A
7.9 cm

680 330
B C

Figure 4.9

Figure 4.11 shows the triangle ABC. Given that ∡ ABC = 680, ∡ ACB = 330, AC = 7.9 cm.
Determine the length of AB.

Solution: : AB = 7.9
Sine Rule sin330 sin680

AB  sin 680 = 7.9  sin 330

AB = 7.9 x sin33
sin68

= 4.64 cm

Example 4.6 :

5.8 cm R
610

S T
7.1 cm

Figure 4.10 Unit 4: Trigonometry

Figure 4.10 shows a triangle RST. Given that ∡SRT = 610, RS = 5.8 cm, ST = 7.1 cm. Find
the value of ∡ STR .

65

UNIT 4 : TRIGONOMETRY

Solution:

RS = ST
Sin T Sin R

5.8 = 7.1
sinT sin61

7.1  sin T = 5.8  sin 61
sinT = sin610 x 5.8
7.1
= 0.8788x5.8
7.1
T = sin−10.7145

∡ STR = 45.60

Practice 4.4 :

1) `Figure 4.8 shows the triangle ABC. Given that AB
= 7cm, BAC = 680, and ACB = 450. Find the
length of BC.

B

7 cm

680 450 C
A

Figure 4.8

Unit 4: Trigonometry

66

UNIT 4 : TRIGONOMETRY

2) Figure 4.9 shows the triangle ABC. Given that BC = 51.3cm C
ABC = 850, and BAC = 810. Find the length of AC. 51.3 cm

A
810

850

B
Figure 4.9

3) Figure 4.10 shows the triangle JKL. Given that JL = 26cm
JKL = 1200, and KJL = 250. Find

a) JLK
K

1200 b) the length of JK
L
250
J

26 cm

Figure 4.10 c) the length of KL

Unit 4: Trigonometry

67

UNIT 4 : TRIGONOMETRY

4) Figure 4.11 shows the triangle RST. Given that RS = 7.9cm, R
ST = 11.9 cm and SRT = 410. Find 410
a) STR

7.9 cm

b) the length of RT. S T
11.9 cm

Figure 4.11

5) Given  ABC = 350, AB = 15 cm, and AC 6) STU is an oblique triangle. Given  T = 470,
= 10 cm. Find  U = 590 and ST = 9 cm. Find the length of
a) ACB SU.

b) BAC

c) The length of BC

Unit 4: Trigonometry

68

UNIT 4 : TRIGONOMETRY

4.4.2 Cosine Rule

B

c a a2 = b2 + c 2 − 2bc cos A
b2 = a2 + c 2 − 2ac cosB
c 2 = a2 + b2 − 2ab cos C

AC
b

Cosine rule can be used when the two sides and one included angle are given.

C

ba

AB

Example 4.7 :

A 15 cm B

680

12 cm

C
Figure 4.12

Figure 4.12 shows the triangle ABC.Given that ∡ABC = 680, AB = 15 cm, BC =12 cm. Find
the length of AC.

Solution: = (15)2 + (12)2 – 2(15)(12) cos 680
= 225 + 144 – 134.86
AC2
AC2 = 234.14
AC = 15.3 cm
AC

Unit 4: Trigonometry

69

UNIT 4 : TRIGONOMETRY
Practice 4.5 :

P 1) Figure 4.12 shows the triangle PQR. Given that
∡ PRQ = 111.40, PR = 20 cm, RQ = 25 cm. Find the
20 cm 111.40 length of PQ.
R 25 cm
Q

Figure 4.12

2) Figure 4.13 shows the triangle PQR. Given that Q
∡ PQR = 57.80, PQ = 13 cm, QR = 10 cm. 57.80
Find the length of PR.
10 cm
13 cm R

P

Figure 4.13

Unit 4: Trigonometry

70

UNIT 4 : TRIGONOMETRY

3) Figure 4.14 shows the triangle ABC. Given that

∡ BAC = 700, AB = 25 cm, AC = 17.2 cm. Find the 25 cm

length of BC. A B
700

17.2 cm

C
Figure 4.14

4) Figure 4.15 shows the triangle MNP. Given that
∡ PMN = 1100, PM = 12 cm, MN = 15 cm.

Find the length of NP.

N P

15 cm 1100 12 cm

M

Figure 4.15

5) Figure 4.16 shows the triangle PQR. Given that ∡ PRQ = 340, PR = 26 cm, QR = 19.9 cm. Find
the length of PQ.

Q
19.9 cm

34

P R
26 cm
Unit 4: Trigonometry
Figure 4.16

71

UNIT 4 : TRIGONOMETRY Area = 1 ab sinC
4.4.3 Area of Triangle 2
The area of oblique triangle is given by :
Area = 1 ac sinB
B 2
a
Area = 1 bc sinA
c 2

AC
b

Example 4.8: Q
13 cm 57.80

10 cm

PR

Figure 4.14

Figure 4.14 shows triangle PQR. Given that PQ = 13 cm, QR = 10 cm and ∡P QR = 57.80.
Find the area of the triangle PQR.

Solution:

The area of the triangle PQR
= 1 x 13 x 10 x sin 57.80
2
= 55.00 cm2

Unit 4: Trigonometry

72

UNIT 4 : TRIGONOMETRY
Practice 4.6 :

Q 1)
11 cm
Figure 4.17 show a triangle PQR. Given that QR =11
cm, PR = 17 cm and ∡ PRQ = 540. Find the area of
triangle PQR.

54

P R
17cm

Figure 4.17

2) Figure 4.18 show a triangle PQR. Given that QR =12.2 cm, Q
PQ = 20.5 cm and ∡ PQR = 650. Find the area of 650
triangle PQR.
12.2 cm
20.5 cm R

P

Figure 4.18

Unit 4: Trigonometry

73

UNIT 4 : TRIGONOMETRY
3) Figure 4.19 shows triangle ABC. Given that ∡ABC = 680, ∡ ACB = 330, and AC = 7.9 cm. Find
a) ∡BAC

b) The length of BC A
c) The area of triangle ABC 7.9 cm

680 33
B C

Figure 4.19

4) Given that ∡ PQR = 720, PQ = 30.5 cm and QR = 21.4 cm. Find
a) The length of PR

b) The area of triangle PQR

Unit 4: Trigonometry

74

UNIT 4 : TRIGONOMETRY Unit 4: Trigonometry

Practice :
1. Given that  ABC = 105 0, Ac = 23 cm and BC = 14 cm. Find
a)  BAC
b) the length of AB
c) the area of triangle ABC

2. Given that  PRQ = 28 0 , QR = 7.6 m, PR = 6.7 m, Find
a) the length of PQ
b) the area of triangle PQR.

3. Given that  ABC = 73 0, AC = 18 cm and BC = 9 cm. Find

a)  BAC
b) the length of AB
c) the area of triangle ABC

4. Given that  PQR = 50 0 , PQ = 5 cm, QR = 12 cm, Find

a) the length of PR
b) the area of triangle PQR

5. Given that  ABC = 45 0, AC = 28 cm and BC = 27 cm. Find

a)  BAC
b) the length of AB
c) the area of triangle ABC

6. Given that  QPR = 52 0 , PQ = 23 m, PR = 25 m, Find
a) the length of RQ
b) the area of triangle PQR.

75