In triangle ABC,if s-a : s-b : s-c = 2 : 3 : | vedclass

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(D) Let $s-a = 2k$,$s-b = 3k$,and $s-c = 4k$. Adding these,we get $3s - (a+b+c) = 9k$. Since $a+b+c = 2s$,we have $3s - 2s = 9k$,so $s = 9k$. Then $a

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$1 : 5$

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(D) Let $s-a = 2k$,$s-b = 3k$,and $s-c = 4k$. Adding these,we get $3s - (a+b+c) = 9k$. Since $a+b+c = 2s$,we have $3s - 2s = 9k$,so $s = 9k$. Then $a = s - 2k = 7k$,$b = s - 3k = 6k$,and $c = s - 4k = 5k$. Using the formula $\cot A = \frac{\cos A}{\sin A} = \frac{b^2+c^2-a^2}{4\Delta}$ and $\cot C = \frac{a^2+b^2-c^2}{4\Delta}$,we have: $\frac{\cot A}{\cot C} = \frac{b^2+c^2-a^2}{a^2+b^2-c^2}$. Substituting the values: $\frac{\cot A}{\cot C} = \frac{(6k)^2 + (5k)^2 - (7k)^2}{(7k)^2 + (6k)^2 - (5k)^2} = \frac{36k^2 + 25k^2 - 49k^2}{49k^2 + 36k^2 - 25k^2} = \frac{12k^2}{60k^2} = \frac{1}{5}$. Thus,$\cot A : \cot C = 1 : 5$.

In $\triangle ABC$,if $s-a : s-b : s-c = 2 : 3 : 4$,then $\cot A : \cot C =$

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