Question

In triangle ABC, $\cot \frac{A}{2}=\frac{b+c}{a}$, then triangle ABC must be
(Note: All symbols used have usual meaning in $△ABC$.]

• A. isosceles
• B. equilateral
• C. right angled
• D. isoceles right angled

Answer 1

The correct option is C right angled

$\cot \left(\frac{A}{2}\right)=\frac{b+c}{a}$

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

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

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

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

$\frac{A}{2}=\frac{B-C}{2}⟹A-B+C=0,-\left(1\right)A+B+C=\Pi -\left(2\right)$

From(1)\& (2)

$2B=\Pi ⟹B=\frac{\Pi }{2}$

$\therefore$ It is a right angled triangle.