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°

⇒ 3x° + 8x° − 17° + 10° = 180°

⇒ 11x° − 7° = 180°

⇒ 11x° – 7° + 7° = 180° + 7                 ...(Adding 7 on both sides.)

⇒ 11x° = 187°

⇒ x° = `187°/11°`

⇒ x° = 17°

Therefore,

∠ACB = (3x − 17)°

= (3 × 17)° − 17°

= (51 − 17)°

= 34°

​∠ACD = (8x + 10)°

= (8 × 17)° + 10°

= (136 + 10)°

= 146°

Now, ∠A + ∠B = ∠ACD                ...(Exterior angle property)

⇒ 2∠A = 146° (∵∠A = ∠B)

⇒ ∠A = `146°/2`

⇒ ∠A = 73°

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