The sides of a triangle are in the ratio $$1:\sqrt 3 :2,$$   then the angles of the triangle are in the ratio

Solution :
Sides are in the Ration $$1:\sqrt 3 :2$$
$$\eqalign{ & {\text{Let }}a = k,b = \sqrt 3 k\,\,{\text{and }}c = 2k \cr & \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{\sqrt 3 }}{2} \cr & \Rightarrow \,\,A = \frac{\pi }{6} \cr & \cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \frac{1}{2} \cr & \Rightarrow \,\,B = \frac{\pi }{3} \cr & \Rightarrow \,\,C = \pi - \left( {A + B} \right) = \frac{\pi }{2} \cr & \Rightarrow \,\,A:B:C = 1:2:3 \cr} $$