With usual notations, if the angles $A, B, C$ of a $ riangle A B C$ are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then $\angle \mathrm{A}=$

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With usual notations, if the angles $A, B, C$ of a $	riangle A B C$ are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then $\angle \mathrm{A}=$

Aicharya · Accepted Answer

Since $A, B, C$ are in A.P., $2 B=A+C \Rightarrow B=60^{\circ}$ We know that, $\frac{\sin B}{b}=\frac{\sin C}{c} \Rightarrow \sin C=\frac{C}{b} 	imes \sin 60^{\circ}$ $	herefore \sin C=\frac{\sqrt{2}}{\sqrt{3}} 	imes \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}$ $	herefore \quad C=45^{\circ} \Rightarrow A=180^{\circ}-\left(60^{\circ}+45^{\circ}ight)=75^{\circ}$

With usual notations, if the angles $A, B, C$ of a $\triangle A B C$ are in A.P. and $b: c=\sqrt{3}:…

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