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(C) Given that the angles $A, B, C$ are in $A$.$P$.,we have $A+C = 2B$. Since $A+B+C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\ci

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) Given that the angles $A, B, C$ are in $A$.$P$.,we have $A+C = 2B$. Since $A+B+C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we have $\frac{b}{c} = \frac{\sin B}{\sin C}$.Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$.This simplifies to $\sin C = \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Thus,$C = 45^{\circ}$.Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

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

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