Mathematics - Properties of Triangle Question with Solution | TestHub

Since $A,B,C$ are in A.P., $2B=A+C$. Also $A+B+C=180^{\circ }$, so $3B=180^{\circ }\Rightarrow B=60^{\circ }$. Using the sine rule, $\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow \frac{\sqrt{3}}{\sin 60^{\circ }}=\frac{\sqrt{2}}{\sin C}$. This gives $\frac{\sqrt{3}}{\sqrt{3}/2}=\frac{\sqrt{2}}{\sin C}\Rightarrow 2=\frac{\sqrt{2}}{\sin C}\Rightarrow \sin C=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$. So $C=45^{\circ }$ (since it's an angle in a triangle). Then $A=180^{\circ }-B-C=180^{\circ }-60^{\circ }-45^{\circ }=75^{\circ }$.