If in a triangle ABC,s(s-a) = (s-b)(s-c),then | vedclass

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(C) Given,$s(s-a) = (s-b)(s-c)$.We know that $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$.Squar

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$\angle A = \frac{\pi}{2}$

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(C) Given,$s(s-a) = (s-b)(s-c)$.We know that $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$.Squaring both,we get $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.Since $s(s-a) = (s-b)(s-c)$,it follows that $\sin^2 \frac{A}{2} = \cos^2 \frac{A}{2}$.Dividing by $\cos^2 \frac{A}{2}$,we get $	an^2 \frac{A}{2} = 1$.Since $\frac{A}{2}$ is an angle of a triangle,$\frac{A}{2} = \frac{\pi}{4}$,which implies $A = \frac{\pi}{2}$.

If in a $\triangle ABC$,$s(s-a) = (s-b)(s-c)$,then

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