HOW TO FIND ANGLES OF A TRIANGLE WITH RATIO

Example 1 :

If the angles of a triangle are in the ratio 5 : 4 : 3, then find the measure of each angle.  

Solution :

From the ratio 5 : 4 : 3, the angles of the triangle are 

5x, 4x and 3x

Sum of the angles of a triangle = 180°

5x + 4x + 3x = 180

Simplify.

12x = 180

Divide each side by 12. 

x = 15

1^st angle = 5(15) = 75°

2^nd angle = 4(15) = 60°

3^rd angle = 3(15) = 45°

Example 2 :

If the angles of a triangle are in the ratio 3 : 4 : 8, then find the measure of each angle.  

Solution :

From the ratio 3 : 4 : 8, the angles of the triangle are 

3x, 4x and 8x

Sum of the angles of a triangle = 180°

3x + 4x + 8x = 180

Simplify. 

15x = 180

Divide each side by 15.

x = 12

1^st angle = 3(12) = 36°

2^nd angle = 4(12) = 48°

3^rd angle = 8(12) = 96°

Example 3 :

In a right triangle ABC, angle A is right angle and the ratio between the angles B and C is 2 : 3. Find the measures of angle B and C.  

Solution :

From the ratio 2 : 3, the angle B and C are 2x and 3x. 

Sum of the angles of a triangle = 180°

m∠A + m∠B + m∠C = 180°

Substitute. 

90 + 2x + 3x = 180

Simplify.

90 + 5x = 180

Subtract 90 from each side.

5x = 90

Divide each side by 5.

x = 18

m∠B = 2(18) = 36°

m∠C = 3(18) = 54°

Example 4 :

In a triangle ABC, measure of ∠A is one of the measure of ∠B and the ratio between the measures of ∠B and ∠C is 2 : 3. Find the measure of each angle.   

Solution :

Given : Measure of angle A is one of the measure of angle B.

∠A = (1/2)∠B

∠A/∠B = 1/2

∠A : ∠B = 1 : 2 ----(1)

Given : Measures of ∠B and ∠C is 2 : 3. Find the measure of each angle.   

∠B : ∠C = 2 : 3 ----(3)

From (1) and (2), ∠A, ∠B and ∠C are in the ratio 1 : 2 : 3.

From the ratio 1 : 2 : 3, the measures ∠A, ∠B and ∠C are 

x, 2x and 3x

Sum of the angles of a triangle = 180°

x + 2x + 3x = 180

Simplify. 

6x = 180

Divide each side by 6.

x = 30

∠A = 30°

∠B = 2(30) = 60°

∠C = 3(30) = 90°

Example 5 :

The ratio of angles in a triangle is 2:3:5 Find the size of the smallest angle.

Solution :

The angles is in the ratio 2 : 3 : 5

Let the angles be 2x, 3x and 5x

Sum of interior angles of triangle = 180

2x + 3x + 5x = 180

10x = 180

x = 180/10

x = 18

• 2x = 2(18) ==> 36
• 3x = 3(18) ==> 54
• 5x = 5(18) ==> 90

So, the required angles are 36, 54 and 90.

Example 6 :

The ratio of three angles in a triangle are 1:2:3. Work out the size of each angle.

Solution :

The angles is in the ratio 1 : 2 : 3

Let the angles be x, 2x and 3x

Sum of interior angles of triangle = 180

1x + 2x + 3x = 180

6x = 180

x = 180/6

x = 30

• x ==> 30
• 2x = 2(30) ==> 60
• 3x = 3(30) ==> 90

So, the required angles are 30, 60 and 90.

Example 7 :

An isosceles triangle has one angle of 52°. Write down the possible sizes of the other two angles in the triangle.

i) Pair 1 …………… and ……………

ii) Pair 2 …………… and ……………

Solution :

i)  Since it is isosceles triangle, let 52 be equal angles.

Let x be the unknown angles.

52 + 52 + x = 180

104 + x = 180

x = 180 - 104

x = 76

The three angles be 52, 52 and 76

ii)  Let the x the equal angles.

x + 52 + x = 180

2x + 52 = 180

2x = 180 - 52

2x = 128

x = 128/2

x = 64

So, the three angles are 64, 64 and 52.

Example 8 :

The ratio of the measures of the three sides of a triangle is 3 : 7 : 5 and its perimeter is 156.8 meters. Find the measure of each side.

Solution :

Let the side lengths be 3x, 7x and 5x

Perimeter of the triangle = 156.8

3x + 7x + 5x = 156.8

15x = 156.8

x = 156.8/15

x = 10.45

3x = 3(10.45) ==> 31.35

7x = 7(10.45) ==> 73.15

5x = 5(10.45) ==> 52.25

So, the three sides are 31.35 m, 73.15 m and 52.25 m.

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