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# Let the angles A, B, C of a triangle ABC be in A. P. and let b : c = √3 : √2. Find the angle A.

7 years ago
As the angles A, B, C of ∆ ABC are in AP
∴ Let A = x – d, B = x, C = x + d
But A + B + C = 180° (∠ Sum prop. of ∆)
∴ x – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60° ∴ ∠B = 60°
Now by sine law in ∆ ABC, we have
b/sin B = c/sin C ⇒ sin B/sin C
⇒ √3/√2 = sin 60°/sin C [using b : c = √3 : √2 and ∠B = 60°]
⇒ √3/√2 = √3/2 sin C ⇒ sin C = 1/√2 = sin 45°
∴ ∠C = 45° ⇒ ∠A = 180° - (∠B + ∠C)
= 180° - (60° + 45°) = 75°
Hello Student,
As the angles A, B, C of ∆ ABC are in AP
∴ Let A = x – d, B = x, C = x + d
But A + B + C = 180° (∠ Sum prop. of ∆)
∴ x – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60° ∴ ∠B = 60°
Now by sine law in ∆ ABC, we have
b/sin B = c/sin C ⇒ sin B/sin C
⇒ √3/√2 = sin 60°/sin C [using b : c = √3 : √2 and ∠B = 60°]
⇒ √3/√2 = √3/2 sin C ⇒ sin C = 1/√2 = sin 45°
∴ ∠C = 45° ⇒ ∠A = 180° - (∠B + ∠C)
= 180° - (60° + 45°) = 75°

Thanks
Rishi Sharma
one year ago
Hello Student,

As the angles A, B, C of ∆ ABC are in AP
∴ Let A = x – d, B = x, C = x + d
But A + B + C = 180° (∠ Sum prop. of ∆)
∴ x – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60° ∴ ∠B = 60°
Now by sine law in ∆ ABC, we have
b/sin B = c/sin C ⇒ sin B/sin C
⇒ √3/√2 = sin 60°/sin C [using b : c = √3 : √2 and ∠B = 60°]
⇒ √3/√2 = √3/2 sin C ⇒ sin C = 1/√2 = sin 45°
∴ ∠C = 45° ⇒ ∠A = 180° - (∠B + ∠C)
= 180° - (60° + 45°) = 75°

Thanks and Regards