The angles \(A, B\) and \(C\) of a triangle \(A B C\) are in AP. If \(\mathrm{b}: \mathrm{c}=\sqrt{3}: \sqrt{2}\), then the angle \(\mathrm{A}\) is

Since A, B, C are in AP.
\(\therefore \quad B=\frac{A+C}{2}\)
\(\Rightarrow \quad \mathrm{B}=90^{\circ}-\frac{\mathrm{B}}{2} \quad\left(\therefore \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}\right)\)
\(\Rightarrow \quad \mathrm{B}=60^{\circ}\)
Using sine rule, \(\quad \frac{\sin B}{b}=\frac{\sin C}{c}\)
\(\therefore \quad \frac{\sin 60^{\circ}}{\sqrt{3}}=\frac{\sin \mathrm{C}}{\sqrt{2}}\)
\(\Rightarrow \quad \frac{\sqrt{3}}{2 \sqrt{3}}=\frac{\sin C}{\sqrt{2}}\)
\(\Rightarrow \quad \sin C=\frac{1}{\sqrt{2}}\)
\[
\begin{aligned}
\Rightarrow & \mathrm{C} &=45^{\circ} \\
\therefore \quad \mathrm{A} &=180^{\circ}-\left(60^{\circ}+45^{\circ}\right) \\
&=75^{\circ}
\end{aligned}
\]