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

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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 180°.

Thus, ∠A +∠B+ ∠C = 180°

∠A + ∠B +3 ∠B = 180°

∠A + 4 ∠B = 180° ..........(ii)

Multiplying 4 to  equation (i) , we get

8∠A – 4∠B = 0 ...........(iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180°

∠A = 20°

Using this in equation (ii), we get

20° + 4∠B = 180°

∠B = 40°

3∠B =∠C

∠C = 3 x 40 = 120°

Therefore, ∠A = 20°

∠B = 40°

∠C = 120°

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