The side BC of ΔABC is produced to D. If ∠ACD = 108° and ∠B = ∠A/2, then ∠A is
We know that the exterior angle is equal to the sum of opposite interior angles.
∠ACD = ∠ABC + ∠BAC
Given, ∠B = $$\frac{\angle A} {2}$$
108° = $$\frac{\angle A} {2}$$ + ∠A
∠A = (108 × 2)/3 = 72°.
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