In a triangle ABC,if r_1 = 2r_2 = 3r_3,then t | vedclass

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(C) We are given $r_1 = 2r_2 = 3r_3 = k$ (let).Then $r_1 = k$,$r_2 = k/2$,and $r_3 = k/3$.The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \

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$5 : 4$

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(C) We are given $r_1 = 2r_2 = 3r_3 = k$ (let).Then $r_1 = k$,$r_2 = k/2$,and $r_3 = k/3$.The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Thus,$s-a = \frac{\Delta}{k}$,$s-b = \frac{2\Delta}{k}$,and $s-c = \frac{3\Delta}{k}$.Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.So,$s = \frac{\Delta}{k} (1 + 2 + 3) = \frac{6\Delta}{k}$.Now,$a = s - (s-a) = \frac{6\Delta}{k} - \frac{\Delta}{k} = \frac{5\Delta}{k}$.And $b = s - (s-b) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$.Therefore,$a : b = \frac{5\Delta}{k} : \frac{4\Delta}{k} = 5 : 4$.

In a $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then the ratio $a : b$ is:

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