In triangle ABC,if a=3, b=4, c=6,then frac{co | vedclass

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(B) We know that $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,where $s = \frac{a+b+c}{2}$ and $\Delta$ is the area of the triangle. Thus,$\cot \frac{A}{

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$\frac{169}{61}$

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(B) We know that $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,where $s = \frac{a+b+c}{2}$ and $\Delta$ is the area of the triangle. Thus,$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \frac{s(s-a) + s(s-b) + s(s-c)}{\Delta} = \frac{s(3s - (a+b+c))}{\Delta} = \frac{s(3s - 2s)}{\Delta} = \frac{s^2}{\Delta} = \frac{(a+b+c)^2}{4\Delta}$. Also,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$. Therefore,the ratio is $\frac{(a+b+c)^2}{a^2+b^2+c^2}$. Substituting $a=3, b=4, c=6$: Ratio $= \frac{(3+4+6)^2}{3^2+4^2+6^2} = \frac{13^2}{9+16+36} = \frac{169}{61}$.

In $\triangle ABC$,if $a=3, b=4, c=6$,then $\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=$

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