Unit 3 - TRIGONOMETRY

DUM10122 TRIGONOMETRY

UNIT MATEMATIK KKTM PASIR MAS 80

DUM10122 TRIGONOMETRY

UNIT 3 :TRIGONOMETRY

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INTRODUCTION

Trigonometry is a branch of mathematics that deals with the relations between the sides
and angles of a triangle.

Nowadays, you also can find the trigonometric ratios of any angles by pressing
appropriate buttons on calculator. Until now we have defined the trigonometric function
only for acute angles. However, many application of trigonometry involves angles that
are not acute. Consequently it is necessary to extend the definition of the six
trigonometry function to general angles and you will learn in this topic. Furthermore in
this chapter, we will derive two new formulae, the sine rule and cosine rule to enable us
to solve oblique triangle quickly. We also can calculate the area of triangles of an oblique
triangles.

LEARNING OUTCOMES
After completing the unit, students should be able to:

1. Convert angle measure from degrees to radians and from radians to
degrees.

2. Form the six trigonometric ratios of a given right-angled.
3. Find the values of :

• trigonometric functions
• inverse trigonometric functions
using calculator.
4. Solve right-angled triangles.
5. Sketch the graph of trigonometric functions.
6. Determine the positive and negative of trigonometric ratios for all
quadrants.
7. Identify the sign of trigonometric ratios for all quadrant.
8. Solve trigonometric equations.
9. Solve oblique triangle using
• Sine Rule
• Cosine Rule.
10. Calculate the area of a given oblique triangle.

UNIT MATEMATIK KKTM PASIR MAS 81

DUM10122 TRIGONOMETRY

3.1 INTRODUCTION TRIGONOMETRY

Angle (Trigonometry)

An angle which has its vertex at the origin, and one side lying on the positive x-
axis. It can have a measure which positive (anticlockwise) or negative
(clockwise) and can be greater than 360°.

Angles are commonly measured in two methods: Degrees & Radians but in
trigonometry radians are the most common.

3.1.1 RELATIONSHIP BETWEEN DEGREE AND RADIAN

180

DEGREE RADIAN

 180


Example 3.1 : Convert degree to radian and radian to degree

1. Convert : c) 4  radian to degree
a) 1350 to radian 9

b) 45° 15’ to radian d) 2.4 radian to degree

Solution: = 45.25o  
a. 135° = 135o   180

180 C. 4  = 4   180
99 
= 2.36 radian = 80o

b. 45°15’ =  45 + 15 0    d. 2.4 radian = 2.4  180
 60  180 

= 137.5o

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= 0.79 radian
3.2 TRIGONOMETRIC RATIOS

Hypotenuse c Hypotenuse c ∡ B
a Opposite a Adjacent

∡A

b b
Adjacent Opposite

Figure 3.1

Hypotenuse : is the longest side of a right angle
triangles and always the side opposite the right
angle.
Opposite side : is opposite the reference angle.
Adjacent side : is next to reference angle.

Try this:
Name the sides of each these right triangles as opposite, adjacent or hypotenuse with
the reference to the state angles.

1. Reference to ∡ Q 2. Reference to ∡ A g

a e
b A
f
Q
c

3. Reference to ∡ R m
R
p
d

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c a

A
b

Figure 3.2

Figure 3.2 shows a right angled triangle where the reference angle is A, the adjacent
side is b, the opposite side is a and the hypotenuse is c.

The six trigonometric functions are defined in Table 3.1.

Function Symbol Definition of function
Sine of angle A Sin A
Cosine of angle A SinA = oppositeside = a
Tangent of angle A Cos A hypotenuse c
Cotangent of angle A
Secant of angle A Tan A CosA = adjacent side = b
Cosecant of angle A hypotenuse c
Cot A = 1
Tan A TanA = oppositeside = a
adjacent side b
Sec A = 1
Cos A CotA = adjacent side = b
1 oppositeside a

Csc A = SecA = hypotenuse = c
Sin A adjacent side b

CscA = hypotenuse = c
oppositeside a

Table 3.1
Formula in Table 3.1 is defined for any angles.

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3.2.1 TRIGONOMETRIC FUNCTIONS

You can find the trigonometric ratios of any angles by pressing appropriate buttons on
your calculator.

Example 3.2 :
Evaluate the trigonometric functions:

a) side PR P 3 R
b) Sin P 4
c) Tan R Q
d) Cos R
e) Sec R
f) Csc P
g) Cot P

Solution:
a) side PR = 42 + 32 = 25 = 5

b) Sin P = 3 = 0.6
5

c) Tan R = 4 = 1.3
3

d) Cos R = 3 = 0.6
5

e) Sec R = 1 = 1 = 1.67
cos R 0.6

f) Csc P = 1 = 1 = 1.67
sin P 0.6

g) Cot P = 4 = 1.33
3

UNIT MATEMATIK KKTM PASIR MAS 100

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Example 3.3 :

Use the calculator to evaluate the trigonometric functions:

a) sin 120 b) cos 2430

c) cot 312.50 d) sec 1300

Solution :

a) sin 120 = 0.2079

b) cos 2430 = -0.4540

c) cot 312.50 = 1 = − 0.9163
Tan 312.5

d) sec 1300 = 1 =−1.5557
Cos130

Example 3.4 :

Determine the value of angle in each of the following given functions.

a) Sin A = 0.7936 d) Sec A = 3.65306

b) Cos A = 0.31236 e) Cot A = 4.8673

c) Tan A = 4.9781 f) Csc A = 2.039

Solution :

a) Sin A = 0.7936
A = Sin-1 0.7956 = 52.520

b) Cos A = 0.3124
A = Cos-1 0.3124 = 71.800

c) Tan A = 4.9781
A = Tan-1 4.9781 = 78.640

d) Sec A = 3.6531 101
1 = 4.8673

CosA
Cos A = 0.2737
A = 74.110

UNIT MATEMATIK KKTM PASIR MAS

DUM10122 TRIGONOMETRY

e) Cot A = 4.8673
1 = 4.8673

Tan A

1 = Tan A
4.8673

Tan A = 0.2063
A = Tan-1 0.2063
A = 11.660

f) Csc A = 2.039
1 = 2.039

Sin A
Sin A = 0.4904
A = 29.370

Example 3.5 :
1. Determine the unknown angles or sides of the right angled triangles.
a) Determine ∡ B

3.9cm B

4.9cm S p
b) Determine: q 72.30

i) ∡ S 16.6 cm
ii) Side p
iii) Side q

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Solution:

a)

Tan B = 3.9 = 0.7959
4.9

B = tan−10.7959

B = 38.510

b) ∡ S = 900 – 72.30
i) = 17.70

ii) cos 72.30 = 16.6
p

p = 16.6 = 16.6
cos72.30 0.3040

= 54.6cm

iii) sin 72.30 = q
54.6

q = 54.6 x sin 72.3
= 52.0cm

UNIT MATEMATIK KKTM PASIR MAS 103

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UNIT EXERCISE 3.2

1. Find the trigonometric ratio of the following angles

(a) sin 360 (b) cot 1240
(c) cos 530 (d) sec 2560
(e) tan 162.20 (f) csc 3130

2. Determine the value of angle for each of following given functions

(a) cos B = 0.37604 (b) sec x = 4.0657
(c) tan D = 0.6945 (d) csc y = 2.9394
(e) sin E = 0.8304 (f) cot θ = 0.1798

3. Find the value of: A
12
(a) side AC B 5C
(b) Sin A
(c) Tan A
(d) Cos C
(e) Sec C
(f) Csc A
(g) Cot A

4. Use calculator to evaluate the trigonometric ratio given in the Table 3.2.

θ Sin θ Cosθ Tan θ

77.20

313.70

202.30

Table 3.2

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5. Solve the right angled triangle

(a)

6.8 cm B

5.8 cm
Figure 3.3
Determine ∡ B from Figure 3.3.

(b)

m 5.3 cm

x y
6.5 cm

Figure 3.4

Determine ∡ x, ∡ y, m from Figure 3.4.

(c)
p

r



180
9.5 cm

Figure 3.5

Determine side r, p and ∡θ from Figure 3.5.

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3.3 GRAPHING TRIGONOMETRIC FUNCTIONS

3.3.1 Graph y = a sin θ for 0  θ < 2π

θ 00 900 1800 2700 3600

(in degree) 0 1   32 2π
2
θ
0 a 0 -a 0
(in radian)

Sin

The graph y = sin θ is shown in Figure 3.6

y
a

0 1   32 2π θ
-a 2

Figure 3.6

The shape of the graph of 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.
This value is also called as amplitude.

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3.3.2 Graph y = a cos θ for 0  θ < 2π

θ 00 900 1800 2700 3600

(in degree) 0 1   32 2π
2
θ

(in radian)

Cos a 0 -a 0 a

The graph of y = cos θ is shown in Figure 3 .7

ay

2Π θ

0 1   32
-a 2

Figure 3.7

The shape of the graph of y = cos x from x = 00 to x = 3600 is repeated for each
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.

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3.3.3 Graph y = a tan θ for 0  θ < 2π

θ 00 900 1800 2700 3600

(in degree) 0 1   32 2π
2
θ

(in radian)

Tan 0 ∞ 0 ∞ 0

The graph of y = tan x is shown in Figure 3.8.
y

0 1   3  2 θ
2 2

Figure 3.8

The shape of the graph of y = tan x from x = 00 to x = 1800 is repeated for each complete
cycle. Hence, the function y = tan x is periodic with a period of 1800.
The function y = tan x does not have any maximum or minimum values.

Example 3.6 :

Sketch the graph of the trigonometric equation given

a) y = 3 sin 2θ 0  θ  2π
b) y = 2 cos 2θ 0 θ π
c) y = sin 2θ 0  θ  360°

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3
Solution: θ
a)
2π
π
2 θ

-3

b) π
y

2

-2

c) 1 θ
-1
180° 360°

UNIT MATEMATIK KKTM PASIR MAS 109

DUM10122 TRIGONOMETRY

Example 3.7 : 360 x
x
Sketch the graph of the trigonometric equation given

a) y = sin 2x 0 °  x  360°

b) y = cos 2x 0 °  x  180°

Solution: y
a) 1

-1 180
b) 180

y
1

-1

UNIT MATEMATIK KKTM PASIR MAS 110

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UNIT EXERCISE 3.3

1. Sketch each of the following trigonometric functions in 0 °  x  360°

(a) y = 2 sin x

(b) y = 1 cos x
2

(c) y = 3 tan x

2. Sketch each of the following trigonometric functions in 0 °  x  2π

(a) y = 2 sin 4x
(b) y = 2 cos 3x
(c) y = 2 sin 3x

3.4 TRIGONOMETRIC EQUATIONS
3.4.1 THE QUADRANT

The Cartesian plane can be divided into 4 quadrants as shown in Figure 3.9.

Quadrants II Quadrants I

Quadrants III Quadrants IV

Figure 3.9

a. Angles in quadrant y
x
In the first quadrant, the angles,  is in 0o ≤  ≤ 90o
commonly called acute angle. This angle is also known is
reference angle,

=α

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In the second quadrant, the angles,  is in 90o ≤  ≤ 180o y
commonly called obtuse angle.
α
 = 180o – α x

In the third quadrant, the angle, is in 180o ≤  ≤ 270o commonly y
called reflex angle. 

 = 180o + α αx

In the fourth quadrant, the angle, is in 270o ≤  ≤ 360o commonly  y
called reflex angle.
x
 = 360o – α α

b. Trigonometric ratio for all quadrants

The sign of the trigonometric ratios can be determine in Figure 4.10 until 4.13.

QUADRANT I

r ● sin θ = y  cosec θ = r
 y r y

x cos θ = x  sec θ = r
r x
Figure 4.10
tan θ = y  cot θ = x
x y

QUADRANT II

● sin θ = y  cosec θ = r
yr r  y

 cos θ = − x sec θ = − r
–x r x

tan θ = − y cot θ = − x
x y

Figure 3.11

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QUADRANT III

–y –x sin θ = − y  cosec θ = − r
●  r  y

r cos θ = − x sec θ = − r
r x

tan θ = y cot θ = x
x y

Figure 3.12

QUADRANT IV

x –y sin θ = − y  cosec θ = − r
 ● r  y

r cos θ = x sec θ = r
r x

tan θ = − y cot θ = − x
x y

Figure 3.13

The four diagrams above can be combined as below:
y

SECOND QUADRANT FIRST QUADRANT
in Positive ll Positive

 x


an Positive os Positive
THIRD QUADRANT FOURTH QUADRANT

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ACTIVITY 3.4.1

Fill-up the Table 4.3 with the sign ( +/– ) of trigonometric.

 Sin  Cos Tan 
45o – +
210o
300o

Table 4.3

You can find the trigonometric ratios of any angles by pressing appropriate
buttons on your calculator.

Example 3.8 :

Use the calculator to evaluate the following:

a) sin 135 b) cos 250 c) cot 45o

Solution: 0.7071
a) sin 135 = – 0.3420
b) cos 250 = 1
c) cot 45o =

ACTIVITY 3.4.2
Use calculator to evaluate the trigonometric ratio given in the Table below.

 Sin  Cos Tan 
45o – 0.8660 0.7071
210o
300o

UNIT MATEMATIK KKTM PASIR MAS 114

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The value of trigonometric function for particular angles (special angles):
30, 45 60

30 30 45 2
22 1 45

3 Figure 3.14 1
60 60

11

The values of the functions sin θ, cos θ and tan θ for the particular angles 30, 45 and
60 are easily obtained without calculator by using the definition of the functions of an
angle in Figure 4.14 and some theorems from plane geometry.

Angle, θ sin θ cos θ tan θ
30
45 1 3 1
60 2 2 3
1 1
2 2 1
3 1
2 2 3

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3.4.2 SOLVING TRIGONOMETRIC EQUATIONS

Trigonometric equation is an equation that involved one or more terms of trigonometric
function. Equation given is a process for getting value that satisfying the equation.

Example 3.9 :
i. sin θ = 1
ii. cos ( θ – 200 ) = 0.7763
iii. sin 2x = 0.2507
iv. sin ( 2x + 40o ) = 0.2

The steps to solve single trigonometric equations are as follows:

1. Determine the quadrants of the angle and should be in based on the
Given trigonometric equation.

2. Find the basic or acute angle using a scientific calculator,  for angle θ.
3. Determine the range of values of the required angles, for example the

range of values of angles 2θ @ 3θ.
4. Determine the values of angles in those quadrants (where is located).

Example 3.10 :
Find the solutions in 0 ≤ θ ≤ 360 of the following equations:

a) tan θ = 0.5
b) cos θ = -0.6428
c) tan 2θ = 1.732

θ

d) sin = 0.7071

2

e) cos (θ - 25) = 0.9848

UNIT MATEMATIK KKTM PASIR MAS 116

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Solution: 26.57
a) tan θ = 0.5
26.57
θ = 26.57
tan θ is positive, so θ are in first and third quadrant

θ = 26.57, 180 + 26.57

= 26.57, 206.57

b) cos θ = -0.6428 50
θ = 50 50

cos θ is negative, so θ are in second and third quadrant
θ = 180 - 50, 180 + 50
θ = 130, 230

c) tan 2θ = 1.732 60
2θ = 60
60
tan 2θ is positive, so 2θ are in first and third quadrant

Given 0 ≤ θ ≤ 360, so 0 ≤ 2θ ≤ 720,

2θ = 60, 180 + 60, 360 + 60, 360 + (180 + 60)

= 60, 240,420,600

θ = 30, 120, 210, 300

θ 45 45

d) sin = 0.7071

2
θ

= 45

2

θθ

sin is positive, so are in first and second quadrant

22
θ

= 45, 180 - 45

2

= 45, 135

θ = 90, 270

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e) cos (θ - 25) = 0.9848 10
(θ - 25) = 10 10

cos (θ - 25) is positive, so (θ - 25) are in first and fourth quadrant
(θ - 25) = 10, 360 - 10
= 10, 350
θ = 10 + 25, 350 + 25
= 35

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UNIT EXERCISE 3.4

1. Find the solutions in 0 ≤ θ ≤ 360 of the following equations.

(a) cosθ = -0.7760 (b) cosecθ = -2

(c) tan 2 θ = 3 θ
(d) sin = 0.6428

2

(e) 5 sin θ = tan θ (f) secθcotθ = 5

2. Solve each of the following trigonometric equations for 0  x  3600

(a) sin x = 0.4233 (d) cos x = 0.3412

(b) tan x = 1.8849 (e) sec x = -2345

(c) cos x = -0.7324 (f) tan x = -2.2755

3. Find the angles between 0° and 360° that satisfy each of the following

trigonometricequations (d) cos 3θ = -0.5473

(a) sin 2θ = 0.5327

(b) tan 2θ = -2.4325 

(e) sin = 0.4453

2

(c) cos 1  = 0.4775 (f) tan 1  = -2.7458
3 2

4. Solve each of the following trigonometric equations for 0O  θ  3600

(a) 2 sin θ = 0.7443 (d) sin 2θ = -0.7569
(b) 3 tan θ = 3.4533
(c) 2 tan 3θ = -0.4357

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3.5 OBLIQUE TRIANGLES
An oblique triangle is a triangle that does not contain a right angle (90o).

Example 3.11:

Figure 3.15 Figure 3.16

The solution of a triangle is defined as a process of finding the length of its three sides
and the values of its three interior angles. For a triangle, the sum of its interior angles is
1800. Figure 3.17 shows an arbitrary triangle.

B

ca

AC
b

Figure 3.17

The angles at the vertices A, B and C will be denoted by A, B and C.
The sides will be denoted by a, b and c.

In this chapter, we will derive two new formulae, the sine rule and cosine rule to enable
us to solve oblique triangle quickly.

UNIT MATEMATIK KKTM PASIR MAS 120

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3.5.1 SINE RULE AND COSINE RULE B
C
In any triangle, the square of any
b side is equal to the sum of the
squares of the other two sides
Ac
minus twice the product of these
In any triangle the two sides multiplied by the
sides are proportional cosine of the included angle

to the sine of other
opposite angles.

Sine Rule can be used when the Cosine Rule can be used when
following values are given: the following values are given:
a. two angles and any one a. two sides and one
included angle is given
side. b. three sides are given.
b. two sides and 1 non-
Cosine Rule
included angle a2 = b2 + c2 - 2bc cos A
(angle opposite one of b2 = a2 + c2 - 2ac cos B
two c2 = a2 + b2 - 2ab cos C
sides given)
or
Sine Rule

a=b=c
sinA sinB sinC

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Example 3.12 :
Figure 3.18 shows the triangle ABC. Given that ∡ ABC = 680, ∡ ACB = 330, AC = 7.9
cm. Determine the length of AB.

A

7.9 cm

680 330 C
B

Solution: Figure 3.18

Sine Rule : c=b
AB sin C sin B
c = 7.9

sin 330 sin 680

7.9x sin33

=

sin680
7.9x0.5446

=

0.9272

= 4.64 cm

Example 3.13 :
Figure 3.19 shows the triangle PQR. Given that ∡ PRQ = 1180, ∡ PQR = 210, PQ = 9.7

cm. Find the length of PR.
P

9.7 cm

1180 Q
R

Figure 3.19

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Solution:

Sine Rule : q=r
sinQ sinR
q = 9.7

sin 210 sin1180

9.7x sin210

q=

sin1180

= 9.7x0.3584 = 3.93cm
0.8829

Example 3.14 :
Figure 3.20 shows the triangle JKL. Given that ∡K = 127.50. ∡J = 250, JL = 26 cm. Find

a) ∡L K j
b) The length of JK l 127.50
250 L
J
26 cm
Solution:
a) ∡ L = 1800 - (250 + 127.50) Figure 3.20

= 1800 - 152.50 123
= 27.50
b) Sine Rule : l = k
sin L sin K

l = 26
sin 27.5 sin127.50

26x sin27.5
l = sin127.50
= 26x0.4617

0.7933

JK = 15.13 cm

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Example 3.15 :
Figure 3.21 shows a triangle RST. Given that ∡R = 61.50, RS = 5.8 cm, ST = 7.1 cm.

Find R
a) ∡T

b) The length of RT 5.8 cm 61.50

S T
7.1 cm
Solution: :
a) Sine Rule Figure 3.21

t =r
sin T sin R

sin T = sin 61.5
5.8 7.1
sin61.50 x5.8

sin T =

7.1
= 0.8788x5.8

7.1

= 0.7179
∡ T = 45.880

b) ∡S = 1800-(45.880 + 61.50)

= 72.620
Sine rule :

 RT = 7.1
sin72.60 sin61.50
sin72.60 x7.1

RT =

sin61.50
= 0.9542x7.1

0.8788

= 7.709 cm

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Example 3.16 :
Figure 3.22 shows the triangle ABC. Given that ∡B = 680, AB = 20.5 cm, BC = 12.2 cm.

Find the length of AC. 20.5 cm B
A 680

12.2 cm

Solution: C
Figure 3.22

b2 = (20.5)2 + (12.2)2 - 2(20.5)(12.2)cos 680
= 408.04 + 148.84 + 187.38
= 381.712

AC = 19.54 cm

Example 3.17 :
Figure 4.25 shows the triangle PQR. Given that ∡R = 111.40, PR = 20 cm, RQ = 25 cm.

Find P
a) Length PQ
b) ∡ Q

20 cm

111.40 Q

R 25 cm

Figure 3.23

Solution: = 202 + 252 - 2(20)(25) cos 111.40
a) r2 = 400 + 625 + 364.87
= 1389.87
PQ = 37.28 cm

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b) sinQ = sin111.4
20cm 37.92cm
sinQ = sin111.4x20
37.9cm

= 0.4913
∡Q = 29.430

Example 3.18 :

Figure 3.24 show the triangle ABC. Given that AB = 6.4 cm, BC = 4.7 cm and AC = 7.8

cm. Find B

a) ∡A

b) ∡C 6.4 cm 4.7 cm

A 7.8 cm C

Solution: Figure 3.24
a)
(6.4)2 + (7.8)2 − (4.7)2

cos A =

2(6.4)(7.8)
40.96 + 60.84 − 22.09

=

99.84
79.71

=

99.84

= 0.7984
∡ A = 37.020

b) sin C = sin 37.20 126
6.4 4.7
6.4 sin37.020

sin∡C =

4.7

sin∡C = 0.8199
∡C = 55.070

UNIT MATEMATIK KKTM PASIR MAS

DUM10122 TRIGONOMETRY

3.5.2 AREA OF TRIANGLE
The area of a triangle in the Figure 4.27 is given by:

B
ca

Ab C
Figure 4.27

A = 1 a x b x sin C
2

= 1 b x c x sin A
2

= 1 a x c x sin B
2

Example 3.19 :
In Figure 3.25 shows Δ PQR. Such that PQ = 13 cm, QR = 10 cm and ∡Q = 57.80. Find

the area of the triangle Δ PQR.

Q

13 cm 57.80 10 cm

P R
Figure 3.25
Solution:
The area of the triangle PQR

= 1 x 13 x 10 sin 57.80
2

= 55.00 cm2

UNIT MATEMATIK KKTM PASIR MAS 127

DUM10122 TRIGONOMETRY

Example 3.20 :
In figure 3.26 shows Δ ABC such that AC = 14 cm, BC = 11 cm , ∡C = 1250. Find

a) AB
b) ∡B
c) The area of the triangle ABC

A

14 cm

1250 B

C 11 cm
Figure 3.26

Solution: = (14)2 + 112 – 2(14)11. cos 1250
a) AB2 = 196 + 121 + 176.7
= 493.7
AB2 = 22.22 cm
AB

b) sin B = sin1250
14 22.22
14x sin1250

sin∡B =

22.22

= 0.5161
∡B = 31.10

c) Area of the triangle ABC

= 1 x 14 x 11 x sin 1250
2

= 63.1 cm2

UNIT MATEMATIK KKTM PASIR MAS 128

DUM10122 TRIGONOMETRY

UNIT EXERCISE 43.5
1. Solve these oblique triangles of ABC given in Table below

ANGLE ( o ) SIDES ( cm )
abc
No. A B C
119
i 126 27 15 21

ii 43 375
58
iii 15 72 228 304

iv 125 32

V 46.3

2. Figure shows the triangle ABC. Given that ∡AB = 810, ∡B = 850,

BC = 51.3 cm. Find AC C
A 51.3 cm
810

850

B

3. Figure shows the triangle PQR.Given that ∡Q = 1240, QR = 16cm, PR =

20cm. Find
(a) ∡P

(b) Length PQ
P

20 cm R
1240
Q 16 cm

UNIT MATEMATIK KKTM PASIR MAS 129

DUM10122 TRIGONOMETRY

4. Figure shows the triangle MNP. Given that ∡P= 240, PN = 36 cm, MN =

19.9 cm.Find: M
(a) ∡N

(b) ∡M 19.9 cm
(c) Length MP

240
NP

36 cm

5. Solve these oblique triangles of ABC given in the Table below :

ANGLE ( 0 ) SIDES ( cm )
No. A B C
abc
i 129
ii 186 179
iii 27.3
iv 51.4 11.3 15.6 12.8
V 35.2
128 152

1.95 1.46

77.3 81.4

6. Figure shows the triangle ABC. Given that ∡A = 700, AB = 25 cm, AC =

17.2 cm. Find: A 25 cm B
(a) BC 700
(b) ∡B

17.2 cm

C

UNIT MATEMATIK KKTM PASIR MAS 130

DUM10122 TRIGONOMETRY

7. Figure shows the triangle PQR. Given that ∡Q = 1070, PQ = 37 cm, QR =
43 cm. Find:
(a) PR
(b) ∡P
(c) The area of the triangle PQR
P

37 cm

107°

Q 43 cm R

8. Figure shows the triangle JKL. Given that JK = 14 cm, KL = 6.9 cm, JL =

9.7 cm. Find: 14 cm
(a) ∡L JK

(b) ∡K 9.7 cm 6.9 cm

(c) The area of the triangle JKL

L

UNIT MATEMATIK KKTM PASIR MAS 131

DUM10122 TRIGONOMETRY

TIPS
When to use sine rule or cosine rule. *(A – Angles, S – Side)

Tips 1: Use sine rule

SSA

AAS

ASA

Tips 2: Use cosine rule

SAS SSS

DICTIONARY

oblique triangle - segitiga yang tidak mempunyai sudut 900
sisi
sides - sudut
garis mencancang
angle - menggantikan
ungkapan
altitude - lakaran
sudut tirus
Substituting - sudut cakah
sudut refleks
Expression -

Sketch -

Acute angle -

Obtuse angle -

Reflex angle -

UNIT MATEMATIK KKTM PASIR MAS 132