in triangleABC, ABC are angles the value of sin(B+C\2)is
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it is-----::::
sin (90-angle A/2)
sin (90-angle A/2)
hrithik92:
the solution
so
okk
a+b+c = 180
b+c=180-a
divide by 2 both side
b+c/2 = 90+ a /2
ohh sorry
90-a/2
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Sin(B+C/2)=sin(90-A/2)[A+B+C=180]
[B+C=180-A]
[B+C/2=90-A/2]
=CosA/2
[B+C=180-A]
[B+C/2=90-A/2]
=CosA/2
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