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

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#### Solution

Given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2 ∠A − ∠B = 0 … (*i*)

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

∠A + ∠B + ∠C = 180°

∠A + ∠B + 3 ∠B = 180°

∠A + 4 ∠B = 180° … (*ii*)

Multiplying equation (*i*) by 4, we obtain

8 ∠A − 4 ∠B = 0 … (*iii*)

Adding equations (*ii*) and (*iii*), we obtain

9 ∠A = 180°

∠A = 20°

From equation (*ii*), we obtain

20° + 4 ∠B = 180°

4 ∠B = 160°

∠B = 40°

∠C = 3 ∠B

= 3 × 40° = 120°

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

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