In a triangle ABC; ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.

As 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)
As 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 gets
⇒8∠A - 4∠B = 0… (iii)
Adding equations (ii) and (iii), we obtain
⇒9∠A = 180°
⇒∠A = 180°/9 = 20°
From equation (ii), we obtain
⇒20° + 4∠B = 180°
⇒4∠B = 160°
⇒∠B = 40°
⇒∠C = 3∠B ⇒3 × 40° = 120°
∴∠A, ∠B, ∠Care 20°, 40°, 120°