Given,

∠C = 3 ∠B = 2(∠B + ∠A)

∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

∠A – ∠B= 0- – – – – – – – – – – –  (i)

We know, the sum of all the interior angles of a triangle is 180^O.

Thus, ∠ A +∠B+ ∠C = 180^O

∠A + ∠B +3 ∠B = 180^O

∠A + 4 ∠B = 180^O– – – – – – – – – – – – – – -(ii)

Multiplying 4 to  equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180^O

∠A = 20^O

Using this in equation (ii), we get

20^O+ 4∠B = 180^O

∠B = 40^O

3∠B =∠C

∠C = 3 x 40 = 120^O

Therefore, ∠A = 20^O

∠B=40^O

∠C = 120^O