In triangle ABC,if cot frac{A}{2} : cot frac{ | vedclass

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(D) Given that,$\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 4 : 3 : 2$. Using the formula $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s

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$5 : 6 : 7$

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(D) Given that,$\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 4 : 3 : 2$. Using the formula $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$,we have: $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}} : \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} : \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = 4 : 3 : 2$. Multiplying each term by $\sqrt{(s-a)(s-b)(s-c)}$,we get: $(s-a) : (s-b) : (s-c) = 4 : 3 : 2$. Let $s-a = 4k$,$s-b = 3k$,and $s-c = 2k$. Adding these equations: $3s - (a+b+c) = 9k$. Since $a+b+c = 2s$,we have $3s - 2s = s = 9k$. Now,$a = s - 4k = 9k - 4k = 5k$. $b = s - 3k = 9k - 3k = 6k$. $c = s - 2k = 9k - 2k = 7k$. Therefore,$a : b : c = 5k : 6k : 7k = 5 : 6 : 7$.

In $\triangle ABC$,if $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 4 : 3 : 2$,then $a : b : c =$

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