Let the angles of a triangle $A B C$ be in A.P. and let $b: c=\sqrt{3}: \sqrt{2}$. Find the angle $A

Question

Let the angles of a triangle $A B C$ be in A.P. and let $b: c=\sqrt{3}: \sqrt{2}$. Find the angle $A in degrees$.

Professor PC Thomas Classes · Answer

$A+B+C=180^{\circ}, 2 B=A+C$\ ( $\because A, B, C$ are in A.P)\ $	herefore \quad 3 B=180^{\circ} \Rightarrow B=60^{\circ}$\ Given $\frac{b}{\sqrt{3}}=\frac{c}{\sqrt{2}}=k \quad 	herefore b=k \sqrt{3}, c=k \sqrt{2}$\ By the sine rule, $\frac{b}{\sin B}=\frac{c}{\sin C}$\ $\Rightarrow \frac{k \sqrt{3}}{\sin 60^{\prime \prime}}=\frac{k \sqrt{2}}{\sin C} \quad 	herefore \quad \sin C=\frac{1}{\sqrt{2}}$\ $C=45^{\circ} \quad\left(\because C=135^{\circ}ight.$ is impossible $)$\ $	herefore \quad A=180^{\circ}-(B+C)$ $$ =180^{\circ}-\left(60^{\circ}+45^{\circ}ight)=75^{\circ} . $$