Question

In $△ABC$, if $\cos C=\frac{\sin A}{2\sin B}$, then show that triangle is right angled triangle.

Answer 1

Consider the diagram of the triangle shown below.

We know that,

$A+B+C=180{}^{\circ }$

$\Rightarrow A=180{}^{\circ }-(B+C)$

$\sin A=\sin (180{}^{\circ }-(B+C))=\sin (B+C)$

Given:

$\cos C=\frac{\sin A}{2\sin B}$

$\sin A=2\sin B\cos C$

$\sin (B+C)=2\sin B\cos C$

$\sin B\cos C+\cos B\sin C=2\sin B\cos C$

$\sin B\cos C-\cos B\sin C=0$

$\sin (B-C)=0$

$B-C=0$

$B=C$

Hence, the triangle is an isosceles triangle because two interior angles are equal to each other.