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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.

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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.

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