Question

The angles of $\Delta ABC$ are in $A.P.$ and $b:c=\sqrt{3}:\sqrt{2}$, then find $\angle A,\angle B,\angle C$.

Answer 1

$\angle A,\angle B$ and $\angle C$ are in A.P.

$\therefore 2B=A+C$

Add B on both sides

$3B=A+B+C={180}^o$ $[\because \angle A+\angle B+\angle C={180}^o]$

$\angle B={60}^o$

By sine rude, In $△ABC$

$\frac{\sin B}{b}=\frac{\sin C}{C}$

$\frac{\sin 60}{\sin c}=\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}\Rightarrow \sin C=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}$

$\sin C=\frac{1}{\sqrt{2}}\Rightarrow C={45}^o$

$\angle A=180-(45+60)$

$\angle A={75}^o$

Thus $\angle A={75}^o,\angle B={60}^o$ and $\angle C={45}^o$.