In triangle ABC,if r_1: r_2 = 7: 8 and r_1: r | vedclass

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(C) Given: $r_1: r_2 = 7: 8$ and $r_1: r_3 = 3: 4$. We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.

Vedclass · Accepted Answer

$13: 14: 15$

Vedclass · Answer

(C) Given: $r_1: r_2 = 7: 8$ and $r_1: r_3 = 3: 4$. We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$. Thus,$\frac{1}{s-a}: \frac{1}{s-b}: \frac{1}{s-c} = r_1: r_2: r_3$. First,find the ratio $r_1: r_2: r_3$: $r_1: r_2 = 7: 8 = 21: 24$ $r_1: r_3 = 3: 4 = 21: 28$ So,$r_1: r_2: r_3 = 21: 24: 28$. Let $s-a = \frac{k}{21}$,$s-b = \frac{k}{24}$,and $s-c = \frac{k}{28}$. Summing these: $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$. So,$s = k(\frac{1}{21} + \frac{1}{24} + \frac{1}{28}) = k(\frac{8+7+6}{168}) = k(\frac{21}{168}) = \frac{k}{8}$. Now,$a = s - (s-a) = k(\frac{1}{8} - \frac{1}{21}) = k(\frac{21-8}{168}) = \frac{13k}{168}$. $b = s - (s-b) = k(\frac{1}{8} - \frac{1}{24}) = k(\frac{3-1}{24}) = \frac{2k}{24} = \frac{14k}{168}$. $c = s - (s-c) = k(\frac{1}{8} - \frac{1}{28}) = k(\frac{7-2}{56}) = \frac{5k}{56} = \frac{15k}{168}$. Therefore,$a: b: c = 13: 14: 15$.

In $\triangle ABC$,if $r_1: r_2 = 7: 8$ and $r_1: r_3 = 3: 4$,then $a: b: c =$

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