If the angles of a triangle are in the ratio $2 : 3 : 7$ ,then the sides are in the ratio 

Solution:

Let ABC be triangle whose angles A, Band Care::-  

the ratio $2 : 3 : 7.$ 

Let $A=2\lambda ,B=3\lambda$ and $\mathrm{AC}=7\lambda$ Then, 

$\begin{array}{l}A+B+C={180}^{\circ }\Rightarrow \lambda =15\\ \therefore A={30}^{\circ },B={45}^{\circ }\text{ and }C={105}^{\circ }\\ \therefore \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \Rightarrow \frac{a}{\sin {30}^{\circ }}=\frac{b}{\sin {45}^{\circ }}=\frac{c}{\sin {105}^{\circ }}\\ \Rightarrow \frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}=\frac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\Rightarrow \frac{a}{\sqrt{2}}=\frac{b}{2}=\frac{c}{\sqrt{3}+1}\end{array}$