Question

# In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

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Solution

## Given that, ∠C = 3∠B = 2(∠A + ∠B) 3∠B = 2(∠A + ∠B) 3∠B = 2∠A + 2∠B ∠B = 2∠A 2 ∠A − ∠B = 0 … (i) We know that the sum of the measures of all angles of a triangle is 180°. Therefore, ∠A + ∠B + ∠C = 180° ∠A + ∠B + 3 ∠B = 180° ∠A + 4 ∠B = 180° … (ii) Multiplying equation (i) by 4, we obtain 8 ∠A − 4 ∠B = 0 … (iii) Adding equations (ii) and (iii), we obtain 9 ∠A = 180° ∠A = 20° From equation (ii), we obtain 20° + 4 ∠B = 180° 4 ∠B = 160° ∠B = 40° ∠C = 3 ∠B = 3 × 40° = 120° Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

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