Question

If A,B and C are interior angles of triangle ABC ,then show that sinB+C/2=CosA/2

Answer 1

Hello Friend....

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The answer of u r question is.....

Ans:

Given,

A,B and C all interior angles of right angle triangle ABC then a + b + c is equals to 80

On dividing the above equation by 2 on both sides,

We get....

$\frac{a}{2 } + \frac{b + c}{2} = 90$

$\frac{b + c}{2} = 90 - \frac{a}{2}$
on taking sin ratio on both side....

$sin \: ( \frac{b + c}{2} ) = sin \:( 90 - \frac{a}{2} )$

$sin \: ( \frac{b + c}{2} ) = cos \frac{a}{2}$

Hence proved....

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Answer 2

Hope it helps..........