Question

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

• A. an isosceles triangle.
• B. an equilateral triangle.
• C. a right angled triangle.
• D. an obtuse angled triangle.

Hint:

Use half-angle and sine rules to simplify triangle identities.

Answer 1

The Correct Option is C

Step 1: Sine Rule
$\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A} = \frac{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}}$.
Step 2: Simplify
Since $\frac{B+C}{2} = 90^\circ - \frac{A}{2}$, $\sin \frac{B+C}{2} = \cos \frac{A}{2}$.
Expression becomes $\frac{\cos \frac{B-C}{2}}{\sin \frac{A}{2}}$.
Step 3: Equate
$\frac{\cos A/2}{\sin A/2} = \frac{\cos (B-C)/2}{\sin A/2} \implies \cos \frac{A}{2} = \cos \frac{B-C}{2}$.
$A = B - C$ or $A = C - B$.
If $A = B - C \implies B = A + C = 90^\circ$.
Final Answer: (C)