In triangle ABC,if a=13, b=14 and cos frac{C} | vedclass

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(C) Given $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$.Using the formula $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we have $\frac{s(s-c

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$s$

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(C) Given $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$.Using the formula $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we have $\frac{s(s-c)}{ab} = \frac{9}{13}$.Substituting $a=13, b=14$,we get $\frac{s(s-c)}{182} = \frac{9}{13}$,so $s(s-c) = 126$.Since $s = \frac{a+b+c}{2} = \frac{27+c}{2}$,then $s-c = \frac{27-c}{2}$.Thus,$\left(\frac{27+c}{2}ight)\left(\frac{27-c}{2}ight) = 126$ $\Rightarrow 729-c^2 = 504$ $\Rightarrow c^2 = 225$ $\Rightarrow c=15$.Then $s = \frac{13+14+15}{2} = 21$.The exradius $r_1$ is given by $r_1 = \frac{\Delta}{s-a}$.Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$.So,$r_1 = \frac{84}{21-13} = \frac{84}{8} = 10.5$.Therefore,$2r_1 = 2 	imes 10.5 = 21 = s$.

In $\triangle ABC$,if $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$,then $2r_1=$

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