If angles of a triangle are in A.P. and b :

Solution:

If angle of a triangle ABC are in A.P 

$\Rightarrow 2 \angle B=\angle A+\angle C$

and $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$

By angle sum property

$\angle A+\angle B+\angle C=\pi$

$\Rightarrow 2 \angle B+\angle B=\pi$

$\Rightarrow \angle B=\frac{\pi}{3}$

also,

Using Sine formula

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

i. e $\frac{b}{\sin B}=\frac{c}{\sin C}$

i. e $\frac{b}{c}=\frac{\sin B}{\sin C}$

i.e $\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin \frac{\pi}{3}}{\sin C} \quad\left(\because \frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}\right.$ given $)$

i. e $\sin C=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}}$

i. e $\sin C=\frac{1}{\sqrt{2}}$

i. e $\sin C=\frac{\pi}{4}$