Solve this.....for me
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In triangle ABC
A+B+C=180
B+C= 180-A
Now taking LHS
Sin(B+C)/2=sin (180-A)/2
=sin (90-A/2)
=cos A/2 RHS proved
I hope it may be helpful
A+B+C=180
B+C= 180-A
Now taking LHS
Sin(B+C)/2=sin (180-A)/2
=sin (90-A/2)
=cos A/2 RHS proved
I hope it may be helpful
Answered by
2
hope this helps you mark me as brain list
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