In a triangle ABC,b:c = sqrt{3}:sqrt{2} and t | vedclass

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(D) Given that in a $	riangle ABC$,angles $A, B, C$ are in $AP$,so $2B = A + C$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which imp

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$75$

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(D) Given that in a $	riangle ABC$,angles $A, B, C$ are in $AP$,so $2B = A + C$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$. Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we get $\frac{b}{c} = \frac{\sin B}{\sin C}$. Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$. This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^{\circ}$. Finally,$\angle A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ}$.

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

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