# In $$\triangle$$ ABC, $$\angle$$C = 3$$\angle$$B = 2($$\angle$$A + $$\angle$$B). Find the three angles of the triangle.

## Solution :

Given,  $$\angle$$C = 2($$\angle$$A + $$\angle$$B)           …….(1)

Adding 2$$\angle$$C on both sides in equation (1), we get

$$\angle$$C + 2$$\angle$$C = 2($$\angle$$A + $$\angle$$B)  + 2$$\angle$$C

$$\implies$$  3$$\angle$$C = 2($$\angle$$A + $$\angle$$B + $$\angle$$C)

Since, $$\angle$$A + $$\angle$$B + $$\angle$$C = 180 degrees

$$\implies$$  $$\angle$$C = $$2\over 3$$ $$\times$$ 180 = 120

Again, $$\angle$$C = 3$$\angle$$B

120 = 3$$\angle$$B  $$\implies$$  $$\angle$$B = 40

But, $$\angle$$A + $$\angle$$B + $$\angle$$C = 180

$$\angle$$A + 40 + 120 = 180

$$\implies$$ $$\angle$$A = 180 – 40 – 120 = 20

Hence, $$\angle$$A = 20, $$\angle$$B = 40, $$\angle$$C = 120