if a b and c are interior angles of a triangle ABC then show that Cos A + B upon 2 is equals to sin C upon 2
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Answered by
8
Answer:
we know that sum of angles of a triangle = 180
A+B+C =180
A+B = 180-C
divide the above equation by 2
= (A+B)/2 = (180-C)/2
(A+B)/2 = (90-C/2)
multiply the above equation by cos on both sides
cos (A+B/2) = cos(90-C/2)
as we know cos (90-x) = sin x
therefore cos A+B/2 = sin C/2
hope it helps.
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0
Answer:
acha question hai excellent
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