Answer:
Option A
Explanation:
In $\triangle$ABC,
$\angle A,\angle B, \angle C $ are in A.P
$\therefore$ 2$\angle B $= $\angle A+ \angle C $.............(i)
and $\angle A+\angle B+ \angle C =180^{0}$..........(ii)
From eqs.(i) and (ii) , B= 60°
$\therefore$ $\frac{b}{\sin B}=\frac{c}{\sin C}$
$\Rightarrow\frac{\sin B}{\sin C}=\frac{b}{c}$
$\Rightarrow\frac{\sin 60^{0}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}$ $[\because$ given $b:c=\sqrt{3}:\sqrt{2}]$
$\Rightarrow \sin C=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times2}=\frac{1}{\sqrt{2}}$
$\therefore$ sin C= sin 45°
$\Rightarrow$ C=45°
$\angle A=2\angle B-\angle C=120^{0}-45^{0}=75^{0}$
