Question

If in a $△ABC,\cos A=\frac{\sin B}{2\sin C}$, prove that it is an isosceles triangle.

Answer 1

We are given that $2\cos A\cos C=\sin B$
or $\sin (A+C)-\sin (A-C)=\sin B....\left(1\right)$
But in a $△,A+C={180}^o-B$ so that
$\sin (A+C)=\sin B$
$\therefore \sin (A-C)=0$ (by (1))
$A-C=0$ or $A=C$
Hence the triangle is isosceles