Question
In a $\triangle\text{ABC},$ if $\cos\text{A}=\frac{\sin\text{B}}{2\sin\text{C}'}$ then show that c = a.

Answer

Let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\cos\text{A}=\frac{\sin\text{B}}{2\sin\text{C}}$
$\Rightarrow\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)=\frac{\text{kb}}{2\text{kc}}$
$\Rightarrow\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{\text{b}}\Big)=\text{b}$
$\Rightarrow\text{b}^2+\text{c}^2-\text{a}^2=\text{b}^2$
$\Rightarrow\text{c}^2=\text{a}^2$
$\Rightarrow\text{c = a}$

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