If the angles \(\mathrm{A}, \mathrm{B}\) and C of a triangle ABC are in the ratio \(2: 3: 7\) respectively, then the sides a, b and c are respectively in the ratio

Angles of triangle ABC are in ratio \(2: 3: 7\)
Let the common multiple be \(x\)
\(\begin{array}{ll}
\therefore & \angle A=2 x, \angle B=3 x, \angle C=7 x \\
\therefore & 2 x+3 x+7 x=180^{\circ}
\end{array}\)
...[Sum of measure of angles of triangle is \(180^{\circ}\) ]
\(\begin{array}{ll}
\therefore & 12 x=180 \\
\therefore & x=15 \\
\therefore & \angle \mathrm{~A}=30^{\circ}, \angle \mathrm{B}=45^{\circ} ; \angle \mathrm{C}=105^{\circ}
\end{array}\)
Now, By sine Rule
\(\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & \frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 105^{\circ}} \\ & \frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}=\frac{c}{\frac{\sqrt{3}+1}{2 \sqrt{2}}} \\ & \frac{a}{\sqrt{2}}=\frac{b}{2}=\frac{c}{\sqrt{3}+1} \\ \therefore \quad & a: b: c=\sqrt{2}: 2: \sqrt{3}+1\end{aligned}\)