Question

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

• A. Isosceles
• B. Equilateral
• C. Right angled
• D. None of these

Answer 1

The correct option is C Right angled

$\cot \frac{A}{2}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}...\left(1\right)$

$\frac{b+c}{a}k\left[\frac{\sin B+\sin C}{k\sin A}\right](\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k)$

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

$\Rightarrow A+B+C=180\Rightarrow \frac{B+C}{2}=90-\frac{A}{2}$

$\Rightarrow \sin \left(\frac{B+C}{2}\right)=\sin (90-\frac{A}{2})=\cos \frac{A}{2}$

$\Rightarrow Now,\frac{b+c}{a}=\frac{\frac{\cos (B-C)}{2}}{\sin \frac{A}{2}}...\left(2\right)$

From $\left(1\right)+\left(2\right)$

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

$\cos \left(\frac{A}{2}\right)-\cos \left(\frac{B-C}{2}\right)=2\times \sin \frac{(B-C+A)}{4}\times \sin \frac{(B-C-A)}{4}=0$

$B-C+A={180}^{0}orB-C-A=0...\left(3\right)$

$A+B+C={180}^0...\left(4\right)$

Solving B-C-A & A+ B+C we get $B=90$

Solving B-C+A & A+ B+C we get $A+B=90$

either $\angle B=90or\angle C=90$

So, a Right Angle Triangle.