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

C = 3B = 2(A + B)

Taking 3B = 2(A + B)

B = 2A

2A -B = 0 ….(i)

We know that the sum of the measures of all angles of a triangle is 180°.

A + B + C =  180⁰

A + B + 3B = 180⁰

A + 4B = 180⁰…….(2)

Multiplying equation (1) by 4, we obtain:

8A – 4B = 0 …….(3)

Adding equations (2) and (3), we get

9A = 180⁰

A = 20⁰

From eq. (2), we get,

20⁰ + 4B = 180⁰

4B = 160⁰

B = 40⁰

C = 3B

3*40⁰ = 120⁰

Hence the measures of A, B and C are 20⁰ ,40⁰ and 120⁰ respectively.