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

step-by-step explanation:

Given, Angle A = Angle B

also,

3x - 17 + 8x + 10 = 180°

→ 11x = 187

→ x = 187/11 = 17

→ x = 17

therefore, Angle ACB 3 ( 17 ) - ( 17 ) = 34°

Angle ACD = 8 ( 17 ) + 10 = 146°

also,

angle A + angle B + Angle ACB = 180°

→ 2angleA + 34° = 180

→ Angle A = 180 - 34 / 2 = 146/2 = 73°

→ Angle A = 73°

Angle A = Angle B = 73°

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