In a triangle ABC,if r_1 : r_2 = 3 : 4 and r_ | vedclass

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(D) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $r_

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$5 : 6 : 7$

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(D) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $r_1 : r_2 = 3 : 4$,we have $\frac{s-b}{s-a} = \frac{3}{4}$,which implies $4s - 4b = 3s - 3a$,or $s = 4b - 3a$.Given $r_2 : r_3 = 2 : 3$,we have $\frac{s-c}{s-b} = \frac{2}{3}$,which implies $3s - 3c = 2s - 2b$,or $s = 3c - 2b$.Equating the two expressions for $s$: $4b - 3a = 3c - 2b \implies 6b = 3a + 3c \implies 2b = a + c$.This shows that $a, b, c$ are in arithmetic progression.Using the property $r_1 : r_2 : r_3 = \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c}$,we have $r_1 : r_2 : r_3 = 3 : 4 : 6$ (since $r_1:r_2=3:4$ and $r_2:r_3=2:3=4:6$).Thus,$s-a = \frac{k}{3}$,$s-b = \frac{k}{4}$,$s-c = \frac{k}{6}$ for some constant $k$.Summing these: $3s - (a+b+c) = k(\frac{1}{3} + \frac{1}{4} + \frac{1}{6}) = k(\frac{4+3+2}{12}) = \frac{9k}{12} = \frac{3k}{4}$.Since $a+b+c = 2s$,we have $3s - 2s = s = \frac{3k}{4}$.Then $a = s - \frac{k}{3} = \frac{3k}{4} - \frac{k}{3} = \frac{5k}{12}$,$b = s - \frac{k}{4} = \frac{3k}{4} - \frac{k}{4} = \frac{2k}{4} = \frac{6k}{12}$,$c = s - \frac{k}{6} = \frac{3k}{4} - \frac{k}{6} = \frac{7k}{12}$.Therefore,$a : b : c = 5 : 6 : 7$.

In a triangle $ABC$,if $r_1 : r_2 = 3 : 4$ and $r_2 : r_3 = 2 : 3$,then $a : b : c =$

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