With usual notations,if the angles A, B, C of | vedclass

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(D) Given that $A, B, C$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we substitute $A + C = 2B$ to get $3B = 180^{\circ}$,wh

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$75$

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(D) Given that $A, B, C$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we substitute $A + C = 2B$ to get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$. Using the Sine Rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,we have $\sin C = \frac{c}{b} \sin B$. Given $b:c = \sqrt{3}:\sqrt{2}$,we have $\frac{c}{b} = \frac{\sqrt{2}}{\sqrt{3}}$. Thus,$\sin C = \frac{\sqrt{2}}{\sqrt{3}} 	imes \sin 60^{\circ} = \frac{\sqrt{2}}{\sqrt{3}} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Therefore,$C = 45^{\circ}$. Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

With usual notations,if the angles $A, B, C$ of a $\triangle ABC$ are in $A$.$P$. and $b:c = \sqrt{3}:\sqrt{2}$,then $\angle A=$ (in $^{\circ}$)

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