In ΔABC if cos C
Question:

In $\Delta A B C$ if $\cos C=\frac{\sin A}{2 \sin B}$, prove that the triangle is isosceles.

Solution:

Let $\Delta A B C$ be any triangle.

Suppose $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

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

$\frac{b^{2}+a^{2}-c^{2}}{2 a b}=\frac{k a}{2 k b} \quad\left(\because \cos C=\frac{b^{2}+a^{2}-c^{2}}{2 a b}\right)$

$\Rightarrow b^{2}+a^{2}-c^{2}=a^{2}$

$\Rightarrow b^{2}-c^{2}=0$

$\Rightarrow(b-c)(b+c)=0$

$\Rightarrow b-c=0$

$\Rightarrow b=c \quad(\because b, c>0)$

Thus, the lengths of two sides of the $\Delta A B C$ are equal.

Hence, $\Delta \mathrm{ABC}$ is an isosceles triangle.