If the angles of a triangle are in the ratio $1 : 2 : 3$, find the ratio of sides opposite to these angles.
Let the angles of a triangle be $A$, $B$ and $C$.
Then, $A + B + C = \pi$ $\;\;\; \cdots \; (1)$
Given: $A : B : C = 1: 2 : 3$
Let $\alpha$ be the multiplicative factor.
Then, $A = \alpha$, $\;$ $B = 2 \alpha$, $C = 3 \alpha$
$\therefore \;$ In view of equation $(1)$ we have,
$\alpha + 2 \alpha + 3 \alpha = \pi$
i.e. $6 \alpha = \pi$ $\implies$ $\alpha = \dfrac{\pi}{6}$
$\therefore \;$ $A = \dfrac{\pi}{6}$, $\;$ $B = 2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}$, $\;$ $C = 3 \times \dfrac{\pi}{6} = \dfrac{\pi}{2}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$ $\;\;\; \cdots \; (2a)$, $b = 2 R \sin B$ $\;\;\; \cdots \; (2b)$, $c = 2 R \sin C$ $\;\;\; \cdots \; (2c)$
Substituting the values of $A$, $B$ and $C$ in equations $(2a)$, $(2b)$ and $(2c)$ respectively, we have,
$a = 2 R \sin \left(\dfrac{\pi}{6}\right) = 2 R \times \dfrac{1}{2} = R$
$b = 2 R \sin \left(\dfrac{\pi}{3}\right) = 2 R \times \dfrac{\sqrt{3}}{2} = \sqrt{3} R$
$c = 2 R \sin \left(\dfrac{\pi}{2}\right) = 2 R \times 1 = 2 R$
$\therefore \;$ $a : b : c = R : \sqrt{3} R : 2 R$
i.e. $\;$ $a : b : c = 1 : \sqrt{3} : 2$