Question

If A, B and C are interior angles of a triangle ABC, then show that
B +C
A
sin
= COS
A
2​

Answer 1

Answer:

Step-by-Step Explanation: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   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)

cosA/2  [sin(90−θ)=cosθ]

Hence sin( B+C /2)=cosA /2

​ proved

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