For the triangle ABC,with usual notations,if | vedclass

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(B) Given that angles $A, B, C$ are in $A.P.$$	herefore A + C = 2B$We know that $A + B + C = 180^{\circ}$Substituting $A + C = 2B$,we get $2B + B = 1

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$\frac{3}{2}, \frac{3 \sqrt{3}}{2}$

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(B) Given that angles $A, B, C$ are in $A.P.$$	herefore A + C = 2B$We know that $A + B + C = 180^{\circ}$Substituting $A + C = 2B$,we get $2B + B = 180^{\circ}$ $\Rightarrow 3B = 180^{\circ}$ $\Rightarrow B = 60^{\circ}$Given $A = 30^{\circ}$,then $C = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 90^{\circ}$Using the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{3}{\sin 90^{\circ}}$$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{3}{1}$$a = 3 	imes \frac{1}{2} = \frac{3}{2}$$b = 3 	imes \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{2}$

For the triangle $ABC$,with usual notations,if the angles $A, B, C$ are in $A.P.$ and $m \angle A = 30^{\circ}, c = 3$,then the values of $a$ and $b$ are respectively

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