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

Answer

Let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}.$ Then, $\sin\text{A = ka},\sin\text{B = kb},\sin\text{C}=\text{kc}$ Now, $\cos\text{C}=\frac{\sin\text{A}}{2\sin\text{B}}$ $2\sin\text{B}\cos\text{C}=\sin\text{A}$ $2\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)\text{kb = ka}$ $\text{a}^2+\text{b}^2-\text{c}^2=\text{a}^2$ $\text{b}^2=\text{c}^2$ $\text{b = c}$ $\triangle\text{ABC}$ is isosceles.

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