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For any triangle ABC prove that Sin B-C/2=B-C/A cos A/

For any triangle ABC prove that Sin B-C/2=B-C/A cos A/

Grade:11

3 Answers

Arun
25763 Points
4 years ago
Let a/sin A = b/sin B = c/sinC = k
Then, 
a = k sinA, b = k sinB, c = k sinC
RHS
b-c/a cos A/2
= (ksinB - ksinC/k sinA )cos A/2
= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2
=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA
= [SinA* sinB-C/2] / sinA
= Sin B-C/2
= LHS
Hence proved
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

Let a/sin A = b/sin B = c/sinC = k
Then,
a = k sinA, b = k sinB, c = k sinC
RHS
b-c/a cos A/2
= (ksinB - ksinC/k sinA )cos A/2
= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2
=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA
= [SinA* sinB-C/2] / sinA
= Sin B-C/2
= LHS
Hence proved

Thanks and Regards
Nishant Kumar
13 Points
9 months ago
R.H.Sb-c /a Cos A/2
          =kSinB - kSinC / kSinA * Cos A/2
         = SinB - SinC / SinA * Cos A/2
=2Cos(B+C)/2 * Sin(B-C)/2 / 2 SinA/2 CosA/2 * CosA/2
Cos( 90 - A/2) Sin(B-C)/2 / Sin A/2
= SinA/2 Sin (B-C)/2 / SinA/2
= Sin (B-C)/2
= R.H.S

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