Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\cot \frac{A}{2}=\frac{b+c}{a}$, then $△ABC$ is?

• A. isosceles
• B. equilateral
• C. right angled
• D. none of these

Answer 1

The correct option is C right angled

Given, $\cot \frac{A}{2}=\frac{b+c}{a}\Rightarrow \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin B+\sin C}{\sin A}$

$\Rightarrow \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)}{2\sin \frac{A}{2}\cos \frac{A}{2}}$

$\Rightarrow \cos \frac{A}{2}=\cos \left(\frac{B-C}{2}\right)\Rightarrow \frac{A}{2}=\frac{B-C}{2}\Rightarrow A=B-C\Rightarrow A+C=B$

but $A+B+C=\pi$

Therefore, $B=\frac{\pi }{2}$

Thus the triangle $ABC$ is a right angled triangle.