In a triangle ABC, sin A - cos B - cos C = 0. What is angle B equal to?
Used Formula:
sin 2θ = 2 sin θ cos θ
cos C + cos D = 2cos((C + D)/2) cos((C − D)/2)
Calculation:
sin A − cos B − cos C = 0
⇒ sin A = cos B + cos C
Using the formula given above
⇒ 2 sin(A/2) cos(A/2) = 2cos((B + C)/2) cos((B − C)/2)
We know that, for ΔABC
∠A + ∠B + ∠C = π
⇒ sin(A/2) cos(A/2) = cos((π − A)/2) cos((B − C)/2)
Since, cos(π/2 − θ) = sin θ
⇒ sin(A/2) cos(A/2) = sin(A/2) cos((B − C)/2)
⇒ cos(A/2) = cos((B − C)/2)
⇒ A/2 = (B − C)/2
⇒ A = B − C
⇒ B = A + C
But ∠A + ∠B + ∠C = π
⇒ B = π − B
⇒ 2B = π
⇒ B = π/2
∴ Angle B is equal to π/2.
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