In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B), then ∠B = ? (a) 20°(b)

1.	In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B), then ∠B = ? (a) 20°(b) 40°(c) 60°(d) 80°
Answer» Correct answer = (b) 40° Let ∠A = x° and ∠B = y° ∴ ∠A = 3∠B = (3y)° Now, ∠A + ∠B + ∠C = 180° ⇒ x + y + 3y = 180 ⇒ x + 4y = 180 ……(i) Also, ∠C = 2(∠A + ∠B) ⇒ 3y = 2(x + y) ⇒ 2x – y = 0 ……..(ii) On multiplying (ii) by 4, we get: 8x – 4y = 0 …….(iii) On adding (i) and (iii) we get: 9x = 180 ⇒ x = 20 On substituting x = 20 in (i), we get: 20 + 4y = 180 ⇒ 4y = (180 – 20) = 160 ⇒ y = 40 ∴ x = 20 and y = 40 ∴ ∠B = y° = 40°

In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B), then ∠B = ? (a) 20°(b) 40°(c) 60°(d) 80°

Answer»

Correct answer = (b) 40°

Let ∠A = x° and ∠B = y°

∴ ∠A = 3∠B = (3y)°

Now, ∠A + ∠B + ∠C = 180°

⇒ x + y + 3y = 180

⇒ x + 4y = 180 ……(i)

Also, ∠C = 2(∠A + ∠B)

⇒ 3y = 2(x + y)

⇒ 2x – y = 0 ……..(ii)

On multiplying (ii) by 4, we get:

8x – 4y = 0 …….(iii)

On adding (i) and (iii) we get:

9x = 180 ⇒ x = 20

On substituting x = 20 in (i), we get:

20 + 4y = 180 ⇒ 4y = (180 – 20) = 160 ⇒ y = 40

∴ x = 20 and y = 40

∴ ∠B = y° = 40°