Question

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If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.

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Answer 1

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➪Given△ABC

➪We know that sum of three angles of a triangle is 180

➪Hence ∠A+∠B+∠C=180°

➪or A+B+C=180°

➪B+C=180° −A

➪Multiply both sides by 1/2

➪1/2 (B+C) = 1/2 (180° − A)

➪1/2 (B + C) = 90° - A/2...(1)

➪Now 1/2 (B+C)

➪Taking sine of this angle

➪sin (B + C /2)

➪[B + C /2 = 90° - A/2]

➪sin(90°-A/2)

➪cos A/2 [sin(90° - θ ) = cos θ]

➪Hence sin( B + C/2 )=cos A/2 proved

❝Hope It Helps❞

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Answer 2

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➪Given△ABC

➪We know that sum of three angles of a triangle is 180

➪Hence ∠A+∠B+∠C=180°

➪or A+B+C=180°

➪B+C=180° −A

➪Multiply both sides by 1/2

➪1/2 (B+C) = 1/2 (180° − A)

➪1/2 (B + C) = 90° - A/2...(1)

➪Now 1/2 (B+C)

➪Taking sine of this angle

➪sin (B + C /2)

➪[B + C /2 = 90° - A/2]

➪sin(90°-A/2)

➪cos A/2 [sin(90° - θ ) = cos θ]

➪Hence sin( B + C/2 )=cos A/2 proved

❝Hope It Helps❞

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