In triangle ABC,find the value of sqrt{frac{r | vedclass

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(B) We know the formulas for the exradii of a triangle: $r = \frac{\Delta}{s-a}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \fra

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$	an(B/2)$

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(B) We know the formulas for the exradii of a triangle: $r = \frac{\Delta}{s-a}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Substituting these into the expression $\sqrt{\frac{r r_2}{r_3 r_1}}$:$\sqrt{\frac{\left(\frac{\Delta}{s}ight) \left(\frac{\Delta}{s-b}ight)}{\left(\frac{\Delta}{s-c}ight) \left(\frac{\Delta}{s-a}ight)}} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Using the half-angle formula $	an^2(B/2) = \frac{(s-a)(s-c)}{s(s-b)}$,we get $\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = 	an(B/2)$.

In $\triangle ABC$,find the value of $\sqrt{\frac{r r_2}{r_3 r_1}}$.

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