TRIGONOMETRY 3 - SINE RULE & COSINE RULE

Chapter 2
TRIGONOMETRY

TRIGONOMETRY

DBM10013 ENGINEERING MATHEMATICHS 1



2.4 APPLY SINE AND COSINE RULES

The Sine Rules

The Cosine Rules

Area of Triangle Using The Formula


A. Sine Rule

• is made up of 6 elements : 3 sides ( a, b, c ) denoted by the small
letters of the opposite vertices and 3 angles ( A, B, C ) denoted by the
capital letters of the vertices.

The sine rule is used to solve triangles when :

o 2 angles and 1 side are given.
o 2 sides and a non-included angle are given.

* A non-included angle refers to an angle that is not contained
between two sides.

EXAMPLE

Solve the triangle ABC if ∠A = 58o, ∠C = 72o21’ and b = 132.8mm

Solution

With using the fact that sum angle in triangle is 180o

∠B = 180o – ( 58o + 72o21’ )
= 49o 39’

Sine law state with:

= =


= 132.8 = 21′
58° 49° 39′ 72°

Solve for a; Solve for c;

= 132.8 21′ = 132.8
58° 49° 39′ 72° 49° 39′

= 132.8 58° = 132.8 72° 21′
49° 39′ 49° 39′

= . c = .

EXAMPLE

Solve the triangle with A = 28.5o, B = 102.2o and a = 45.5cm.

Solution

With information that given, we draw and label triangle ABC as show at
figure. We find

Solution ∠C = 180o – (28.5o + 102.2o)
= 49.3 o

Solve for b;

45.5 = =
28.5° 102.2° 49.3°

= 45.5
102.2° 28.5°

= 45.5 102.2°
28.5°

= .

Solve for c;

= 45.5
49.3° 28.5°

= 45.5 49.3°
28.5°

= .

EXERCISE



EXERCISE

B. Cosine Rule

The cosine rule is used to solve triangles when :
o 3 sides are given
o 2 sides and the included angle are given

*The included angle refers to the angle contained between two given sides.

EXAMPLE

In , = 7.5 , = 6.4 and = 5.8 . Solve the triangle by
finding the angles.

Solution

Using the cosine rule,

2 = 2 + 2 − 2

= 2+ 2− 2
2

= 7.52+6.42−5.82
2 7.5 6.4

∠ = −1 0.6622

= . °

Solution

Now, using the sine rule,

=

sine sine

6.4 = 5.8
sine sine 48.53°

sine = 6.4 sine 48.53° ∠ = 180° − ∠ − ∠
= 180° − 48.53° − 55.77°
5.8 = . °

∠ = sine−10.8268

= . °

Thus, the angles of the triangle are ∠ = . °, ∠ = . ° and ∠ = . °

EXAMPLE

In , = 1.8 , = 2.3 , ∠ = 46°. Solve the triangle.

2.3

Solution

Using the cosine rule,

2 = 2 + 2 − 2
2 = 1. 82 + 2. 32 − 2 1.8 2.3 46°

= 1. 82 + 2. 32 − 2 1.8 2.3 46°
= .

Now, using the sine rule,

=

sine sine

1.8 = 1.667
sine 46°
sine

sine = 1.8 sine 46° ∠ = 180° − ∠ − ∠
= 180° − 46° − 50.96°
1.667 = . °

∠ = sine−10.7767

= . °

Thus, the unknown side and angles are = . , ∠ = . °, and ∠ = . °

EXERCISE

1.

EXERCISE

SINE SIDE ANGLE
RULE 2 1

1 2

COSINE 3 NONE
RULE 2 1

2.3

Area of Triangle

The simplest way to find the area of a Area = × base × perperdicular height
triangle is to use the formula

If the perpendicular height is not
known, we can use the formula

EXAMPLE

Find the area of triangle PQR, given that p = 30 cm, q = 40 cm and R = 79°

Solution

EXERCISE

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