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

Solution: 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 180O. Thus, ∠ A +∠B+ ∠C = 180O ∠A + ∠B +3 ∠B = 180O ∠A + 4 ∠B = 180O– – – – – – – – – – – – – – -(ii) Multiplying 4 to equation (i) , we get 8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii) Adding equations (iii) and (ii) we get 9 ∠A = 180O ∠A = 20O Using this in equation (ii), we get 20O+ 4∠B = 180O ∠B = 40O 3∠B =∠C ∠C = 3 x 40 = 120O Therefore, ∠A = 20O ∠B=40O ∠C = 120O