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°