In a , if = 2 , prove that the triangle is isosceles.

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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