In a triangle ABC, if (1/2)∠A + ∠B + ∠C = 120º, find the value of (1/4)∠A.

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Let ∠A = x, ∠B = y, ∠C = z. In any triangle, the sum of angles is x + y + z = 180º. Given: (1/2)x + y + z = 120º. Subtract the second equation from the first: (x + y + z) - [(1/2)x + y + z] = 180º - 120º ⇒ x - (1/2)x = 60º ⇒ (1/2)x = 60º ⇒ x = 120º. Now, (1/4)∠A = (1/4) × 120º = 30º. Thus, the answer is 30º.

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