Question

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

Answer 1

Let $∆ABC$ 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

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

^{$$\begin{gathered} \Rightarrow b^2+a^2-c^2=a^2 \\ \Rightarrow b^2-c^2=0 \\ \Rightarrow \left(b-c\right)\left(b+c\right)=0 \\ \Rightarrow b-c=0 \\ \Rightarrow b=c\left(\because b,c>0\right) \end{gathered}$$}

Thus, the lengths of two sides of the $∆ABC$ are equal.

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