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