If in a triangle the angles are in A.P. and b | vedclass

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(D) Let the angles of the triangle be $A-d, A, A+d$. Since the sum of angles in a triangle is $180^o$,we have $(A-d) + A + (A+d) = 180^o$,which implie

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$75$

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(D) Let the angles of the triangle be $A-d, A, A+d$. Since the sum of angles in a triangle is $180^o$,we have $(A-d) + A + (A+d) = 180^o$,which implies $3A = 180^o$,so $A = 60^o$. Given $b:c = \sin B : \sin C = \sqrt{3} : \sqrt{2}$. Since $B = A = 60^o$ is not necessarily true (the angles are $A-d, A, A+d$),let the angles be $B-d, B, B+d$. Then $B = 60^o$. Using the sine rule,$\frac{\sin B}{\sin C} = \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$. Substituting $B = 60^o$,we get $\frac{\sin 60^o}{\sin C} = \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$. This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^o$. Since $A+B+C = 180^o$,we have $A + 60^o + 45^o = 180^o$,which gives $A = 75^o$.

If in a triangle the angles are in $A.P.$ and $b:c = \sqrt{3}:\sqrt{2}$,then $\angle A$ is equal to .....$^o$.

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