In a triangle ABC with usual notations,if cot | vedclass

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(C) Using the half-angle formula for $\cot \frac{A}{2}$,we have $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$.Given $\cot \frac{A}{2} = \frac{

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a right angled triangle.

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(C) Using the half-angle formula for $\cot \frac{A}{2}$,we have $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$.Given $\cot \frac{A}{2} = \frac{b+c}{a}$,we substitute the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.Then $\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A} = \frac{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}}$.Since $A+B+C = \pi$,$\sin \frac{B+C}{2} = \cos \frac{A}{2}$.Thus,$\frac{b+c}{a} = \frac{\cos \frac{A}{2} \cos \frac{B-C}{2}}{\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\cos \frac{B-C}{2}}{\sin \frac{A}{2}}$.Equating this to $\cot \frac{A}{2} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$,we get $\cos \frac{B-C}{2} = \cos \frac{A}{2}$.Since $\frac{A}{2} = \frac{\pi}{2} - \frac{B+C}{2}$,we have $\cos \frac{B-C}{2} = \sin \frac{B+C}{2}$.This implies $\cos \frac{B-C}{2} = \cos (\frac{\pi}{2} - \frac{B+C}{2}) = \cos (\frac{A}{2})$.This leads to $\frac{B-C}{2} = \frac{\pi}{2} - \frac{B+C}{2}$ or $\frac{B-C}{2} = -(\frac{\pi}{2} - \frac{B+C}{2})$.Solving $\frac{B-C}{2} = \frac{\pi}{2} - \frac{B+C}{2}$ gives $B = \frac{\pi}{2}$,which means the triangle is a right-angled triangle.

In a triangle $ABC$ with usual notations,if $\cot \frac{A}{2} = \frac{b+c}{a}$,then the triangle $ABC$ is

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