Question

In \[∆ ABC \text{ if } \cos C = \frac{\sin A}{2 \sin B}\] prove that the triangle is isosceles.  

Answer 1

Let \[∆ ABC\] be any triangle. 

Suppose \[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k\] 

If  \[\cos C = \frac{\sin A}{2\sin B}\]  then 

\[\frac{b^2 + a^2 - c^2}{2ab} = \frac{ka}{2kb} \left( \because \cos C = \frac{b^2 + a^2 - c^2}{2ab} \right)\] 

\[\Rightarrow b^2 + a^2 - c^2 = a^2 \]

\[ \Rightarrow b^2 - c^2 = 0\]

\[ \Rightarrow \left( b - c \right)\left( b + c \right) = 0\]

\[ \Rightarrow b - c = 0\]

\[ \Rightarrow b = c \left( \because b, c > 0 \right)\]

Thus, the lengths of two sides of the  \[∆ ABC\] are equal. 

Hence, \[∆ ABC\] is an isosceles triangle.