Question

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.

Answer 1

Given:
∠ACB = (3x $-$ 17)^∘
∠ACD = (8x + 10)^∘
Now, ∠ACB + ∠ACD = 180^∘ (Linear Pair angles)
⇒ 3x − 17 + 8x + 10 = 180
⇒ 11x = 187
⇒ x = 17
Therefore,
∠ACB = (3x $-$ 17)^∘
= (51 $-$ 17)^∘
= 34^∘
​∠ACD = (8x + 10)^∘
= (136 + 10)^∘
= 146^∘
Now, ∠A + ∠B = ∠ACD (Exterior angle property)
⇒ 2∠A = 146^∘ (∵∠A = ∠B)
⇒ ∠A = 73^∘
Hence, the measures of ∠ACB, ∠ACD, ∠A and ∠B are 146^∘, 34^∘, 73^∘ and 73^∘ respectively.