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

#### Question

#### Answer

Using the Law of Cosines

\cos C = \dfrac{a^2+b^2-c^2}{2ab}

Using the Law of sines

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=2R

\dfrac{\sin A}{2\sin B}=\dfrac{a}{2b}

So the given condition is converted to

\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{a}{2b}

Simplifying the equation yields

b=c

Now we have verified the triangle is an isosceles triangle.