The angles of $\triangle ABC$ are in arithmetic progression. If $b : c = \sqrt{3} : \sqrt{2}$, find $\angle A$.
Let the angles of $\triangle ABC$ be $A, \; B, \; C$.
Given: $\;$ $\angle A, \; \angle B, \; \angle C$ $\;$ are in arithmetic progression.
$\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$
$\because \;$ $A, \; B, \; C$ are the angles of $\triangle ABC$,
$A + B + C = \pi$
i.e. $\;$ $2 B + B = \pi$ $\;\;\;$ [By equation $(1)$]
$\implies$ $B = \dfrac{\pi}{3}$
By law of sines, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\therefore \;$ We have $\;$ $\dfrac{b}{c} = \dfrac{\sin B}{\sin C}$
Given: $\;$ $\dfrac{b}{c} = \dfrac{\sqrt{3}}{\sqrt{2}}$
$\therefore \;$ $\dfrac{\sin B}{\sin C} = \dfrac{\sqrt{3}}{\sqrt{2}}$ $\;\;\; \cdots \; (2)$
Substituting the value of $B$ in equation $(2)$, we have,
$\sin C = \dfrac{\sqrt{2}}{\sqrt{3}} \times \sin \left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{2} = \dfrac{1}{\sqrt{2}}$
$\implies$ $C = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$
$\therefore \;$ $A = \pi - \left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right) = \dfrac{5 \pi}{12} = 75^{\circ}$