In triangle ABC, with usual notations, if the | vedclass

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(C) Given $a:b:c = 18:17:7$. Let $a = 18k, b = 17k, c = 7k$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{18k+17k+7k}{2} = 21k$.Using the formula $\

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$3 : 4 : 14$

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(C) Given $a:b:c = 18:17:7$. Let $a = 18k, b = 17k, c = 7k$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{18k+17k+7k}{2} = 21k$.Using the formula $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$, we calculate:$\cot \frac{A}{2} = \sqrt{\frac{21k(21k-18k)}{(21k-17k)(21k-7k)}} = \sqrt{\frac{21k \cdot 3k}{4k \cdot 14k}} = \sqrt{\frac{63}{56}} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}}$.$\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{21k(21k-17k)}{(21k-18k)(21k-7k)}} = \sqrt{\frac{21k \cdot 4k}{3k \cdot 14k}} = \sqrt{\frac{84}{42}} = \sqrt{2}$.$\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{21k(21k-7k)}{(21k-18k)(21k-17k)}} = \sqrt{\frac{21k \cdot 14k}{3k \cdot 4k}} = \sqrt{\frac{294}{12}} = \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}}$.Ratio $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = \frac{3}{2\sqrt{2}} : \sqrt{2} : \frac{7}{\sqrt{2}} = 3 : 4 : 14$.

In $\triangle ABC$, with usual notations, if the sides $a, b$ and $c$ are in the ratio $18 : 17 : 7$, then $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = $

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