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