If in a triangle $A B C$, with usual notations, the angles are in A.P. and $\mathrm{b}: \mathrm{c}=\sqrt{3}:…
If in a triangle $A B C$, with usual notations, the angles are in A.P. and $\mathrm{b}: \mathrm{c}=\sqrt{3}: \sqrt{2}$, then angle. $\mathrm{A}=$
$30^{\circ}$
$60^{\circ}$
$75^{\circ}$
$45^{\circ}$
Solution
Since the angles are in A.P., therefore $B=60^{\circ}$ By sine rule,
$\begin{aligned}
& \frac{b}{c}=\frac{\sin B}{\sin C} \Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{2 \sin C} \Rightarrow C=45^{\circ} \\
\therefore \quad & A=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}
\end{aligned}$