Question 1.

∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

Solution:

Let the measures of ∠A be x°.

m∠A = m∠B = x°

∠ACD is the exterior angle of ∆ABC

∴ m∠ACD = m∠A + m∠B

∴ 140 = x + x

∴ 140 = 2x

∴ 2x = 140

∴ x =

= 70

∴ The measures of the angles ∠A and ∠B is 70° each.

Question 2.

Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Solution:

m∠EOD = m∠AOB = 8y ….(vertically opposite angles)

∠FOL, ∠EOD and ∠COD form a straight angle.

∴ m∠FOE + m∠EOD + m∠COD = 180°

∴ 4y + 8y + 6y = 180

∴ 18y = 180

∴ y = 180/18

∴ y = 10

m∠EOD = 8y = 8 x 10 = 80°

m∠AOF = m∠COD ….(Vertically opposite angles)

= 6y = 6 x 10 = 60°

m∠BOC = m∠FOE ….(Vertically opposite angles)

= 4y = 4 x 10 = 40°

∴ The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

Question 3.

In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

Solution:

Let the measure of ∠A be y°. A

∴ m∠A = m∠B = y°

∠ACB and ∠ACD form a pair of linear angles.

∴ m∠ACB + m∠ACD = 180°

∴ (3x – 17) + (8x + 10) = 180

∴ 3x + 8x – 17 + 10 = 180

∴ 11x – 7 = 180

∴ 11x – 7 + 7 = 180 + 7 …(Adding 7 on both sides.)

∴ 11x = 187

∴ x = 187/11 = 17

m∠ACB = 3x – 17 = (3 x 17) – 17 = 51 – 17 = 34°

m∠ACD = 8x + 10 = 8 x 17 + 10 = 136 + 10 = 146°

Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.

∴ m∠ACD = m∠A + m∠B

∴ 146 = y + y

∴ 146 = 2y

∴ 2y = 146

∴ y = 146/2 = 73

∴ The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.

**Intext Questions and Activities**

Question 1.

Use straws or sticks to make all the kinds of angles that you have learnt about. (Textbook pg. no. 29)

Solution:

(Student should attempt the activity on their own)

Question 2.

Observe the table given below and draw your conclusions (Textbook pg. no. 31)

Solution:

i. 180°

ii. 360°

iii. 540°

iv. 720°

v. 180° x 5 = 900°

vi. , 180° x 6 = 1080°