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If $\cot \frac{A}{2} = \frac{b + c}{a},$ then the $△ A B C,$

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a

isosceles

b

equilateral

c

right angled

d

none of these

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detailed solution

Correct option is C

We have,

$\begin{array}{l} \cot \frac{A}{2} = \frac{b}{a} + \frac{c}{a} \\ \Rightarrow \frac{\cos A / 2}{\sin A / 2} = \frac{\sin B + \sin C}{\sin A} \\ \Rightarrow \frac{\cos A / 2}{\sin A / 2} = \frac{2 \sin (\frac{B + C}{2}) \cos (\frac{B - C}{2})}{2 \sin \frac{A}{2} \cos \frac{A}{2}} \end{array}$

$\begin{array}{l} \Rightarrow \cos \frac{A}{2} = \cos (\frac{B - C}{2}) \\ \Rightarrow \frac{A}{2} = \frac{B - C}{2} \\ \Rightarrow A = B - C \\ \Rightarrow A + C = B \Rightarrow A + B + C = 2 B \Rightarrow π = 2 B \Rightarrow B = π / 2 \end{array}$

Hence, $∆ A B C$ is right angled.

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