In a $$\Delta \,ABC,$$   if $$\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c},$$      and the side $$a = 2,$$  then are $$a$$ of the triangle is

Solution :
$$\eqalign{ & \frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c} \cr & \Rightarrow \frac{{\cos A}}{{2R\sin A}} = \frac{{\cos B}}{{2R\sin B}} = \frac{{\cos C}}{{2R\sin C}} \cr & \Rightarrow \cot A = \cot B = \cot C \cr & \Rightarrow A = B = C = {60^ \circ } \cr & \Rightarrow \Delta \,ABC{\text{ is equilateral}} \cr & {\text{Hence}},\,\,\Delta = \frac{{\sqrt 3 }}{4}{a^2} = \sqrt 3 . \cr} $$