In triangle ABC,if r_1 = 2r_2 = 3r_3,then b : | vedclass

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(A) Given,$r_1 = 2r_2 = 3r_3 = k$ (let). Since $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have: $s-a = \f

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$4 : 3$

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(A) Given,$r_1 = 2r_2 = 3r_3 = k$ (let). Since $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have: $s-a = \frac{\Delta}{k}$,$s-b = \frac{\Delta}{2k}$,and $s-c = \frac{\Delta}{3k}$. Summing these: $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$. Thus,$s = \Delta(\frac{1}{k} + \frac{1}{2k} + \frac{1}{3k}) = \Delta(\frac{6+3+2}{6k}) = \frac{11\Delta}{6k}$. Now,$b = s - (s-b) = \frac{11\Delta}{6k} - \frac{\Delta}{2k} = \frac{11\Delta - 3\Delta}{6k} = \frac{8\Delta}{6k} = \frac{4\Delta}{3k}$. And $c = s - (s-c) = \frac{11\Delta}{6k} - \frac{\Delta}{3k} = \frac{11\Delta - 2\Delta}{6k} = \frac{9\Delta}{6k} = \frac{3\Delta}{2k}$. Therefore,$b : c = \frac{4\Delta}{3k} : \frac{3\Delta}{2k} = \frac{4}{3} : \frac{3}{2} = 8 : 9$. Wait,re-evaluating the ratio: $b : c = 4/3 : 3/2 = 8 : 9$. Let us re-check the initial condition: $r_1 = 2r_2 = 3r_3$. $s-a = \frac{\Delta}{r_1}$,$s-b = \frac{\Delta}{r_2} = \frac{2\Delta}{r_1}$,$s-c = \frac{\Delta}{r_3} = \frac{3\Delta}{r_1}$. $3s - (a+b+c) = s = \Delta(\frac{1}{r_1} + \frac{2}{r_1} + \frac{3}{r_1}) = \frac{6\Delta}{r_1}$. $b = s - (s-b) = \frac{6\Delta}{r_1} - \frac{2\Delta}{r_1} = \frac{4\Delta}{r_1}$. $c = s - (s-c) = \frac{6\Delta}{r_1} - \frac{3\Delta}{r_1} = \frac{3\Delta}{r_1}$. Thus,$b : c = 4 : 3$.

In $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then $b : c$ equals

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