Question

if sin (A+B-C) 1/2 , cos (B+C-A) = 1/√2. Find A , B and C





Plzz answer if you know​

Answer 1

Answer:

Given,

A + B + C = 180° ⟶ (1)

sin (A + B − C) = 1/2 ⟶ (2)

cos(B + C − A) = 1/√2 ⟶(3)

Now from (1)

A + B = Π − C⟶(4). (Here Π is some angle or 'x')

∴ Using (4) we can rewrite (2) as

sin (Π − 2C) = 1/2

⇒sin 2C = 1/2. (∵sin (Π − θ) = sinθ)

⇒sin 2C = 1/2

∴ 2C = 30° or 150° (∵ sin 30° = 1/2)

⇒C = 15° or 75° (∵ sin 150° = 1/2)

Again from (1)

B + C = Π − A

∴ cos (B + C − A) = 1/√2

⇒cos (Π − 2A) = 1/√2

⇒−cos 2A = 1/√2 (∵ cos(Π−θ)=−cosθ)

⇒cos 2A = -1/√2

∴ 2A = 135°. (∵ cos135° = -1/√2)

⇒A = 67.5°

Case 1 : For C = 15°

Case 2 : For C = 75°

A = 67.5°

∴ B = 180° − (15° + 67.5°) B = 180° −(67.5° + 75°)

= 180° − 82.5° = 97.5° = 37.5°

Since ΔABC is acute angled, thereby n none of the angles can be more than 90°

Hence ∠A = 67.5° , ∠B = 37.5 , ∠C = 75°

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