With usual notations,if in triangle ABC,s is | vedclass

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(C) We know the half-angle formulas for a triangle:$\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$ and $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$.Given the

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a right-angled triangle

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(C) We know the half-angle formulas for a triangle:$\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$ and $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$.Given the condition $(s-a)(s-b) = s(s-c)$,we can substitute these into the expressions:$ab \sin^2 \frac{C}{2} = ab \cos^2 \frac{C}{2}$.Dividing both sides by $ab \cos^2 \frac{C}{2}$ (assuming $ab 
eq 0$ and $\cos \frac{C}{2} 
eq 0$):$	an^2 \frac{C}{2} = 1$.Since $\frac{C}{2}$ must be an acute angle in a triangle,$	an \frac{C}{2} = 1$ implies $\frac{C}{2} = 45^{\circ}$.Therefore,$C = 90^{\circ}$.Thus,$	riangle ABC$ is a right-angled triangle.

With usual notations,if in $\triangle ABC$,$s$ is the semi-perimeter and $(s-a)(s-b)=s(s-c)$,then $\triangle ABC$ is

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