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