In triangle ABC,if A, B, C are in arithmetic | vedclass

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(C) Given that $A, B, C$ are in arithmetic progression,so $A+C = 2B$. Since $A+B+C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\c

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(C) Given that $A, B, C$ are in arithmetic progression,so $A+C = 2B$. Since $A+B+C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.Given $r_1 r_2 = r_3 r$,we use the formulas $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Substituting these,we get $\frac{\Delta^2}{(s-a)(s-b)} = \frac{\Delta^2}{s(s-c)}$,which simplifies to $(s-a)(s-b) = s(s-c)$.This is equivalent to $	an^2 \frac{C}{2} = 1$,so $\frac{C}{2} = 45^{\circ}$,which means $C = 90^{\circ}$.Since $B = 60^{\circ}$ and $C = 90^{\circ}$,then $A = 30^{\circ}$.The area $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} (2R \sin A)(2R \sin B) \sin 90^{\circ} = 2R^2 \sin 30^{\circ} \sin 60^{\circ} = 2R^2 (\frac{1}{2}) (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} R^2$.Given $\Delta = \frac{\sqrt{3}}{2}$,we have $\frac{\sqrt{3}}{2} R^2 = \frac{\sqrt{3}}{2}$,so $R^2 = 1$,which gives $R = 1$.

In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,$\Delta = \frac{\sqrt{3}}{2}$ and $r_1 r_2 = r_3 r$,then $R =$

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