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

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(A) Given $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$. Using the formula $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,we have:

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$8 : 6 : 5$

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(A) Given $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$. Using the formula $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,we have: $\frac{s(s-a)}{\Delta} : \frac{s(s-b)}{\Delta} : \frac{s(s-c)}{\Delta} = 3 : 7 : 9$. Multiplying by $\frac{\Delta}{s}$,we get $(s-a) : (s-b) : (s-c) = 3 : 7 : 9$. Let $s-a = 3k$,$s-b = 7k$,and $s-c = 9k$. Adding these,we get $3s - (a+b+c) = 19k$. Since $a+b+c = 2s$,we have $3s - 2s = s = 19k$. Now,$a = s - 3k = 19k - 3k = 16k$. $b = s - 7k = 19k - 7k = 12k$. $c = s - 9k = 19k - 9k = 10k$. Thus,$a : b : c = 16k : 12k : 10k = 16 : 12 : 10 = 8 : 6 : 5$.

In $\triangle ABC$,if $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$

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