In triangle ABC,with usual notations,if frac{ | vedclass

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(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$. $b+c=11k$ $(i)$,$c+a=12k$ $(ii)$,$a+b=13k$ $(iii)$. Adding $(i), (ii), (iii)$,we get $2(a+b+

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$\frac{51}{35}$

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(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$. $b+c=11k$ $(i)$,$c+a=12k$ $(ii)$,$a+b=13k$ $(iii)$. Adding $(i), (ii), (iii)$,we get $2(a+b+c)=36k$,so $a+b+c=18k$ $(iv)$. From $(iv)$,$a=(a+b+c)-(b+c)=18k-11k=7k$. $b=(a+b+c)-(c+a)=18k-12k=6k$. $c=(a+b+c)-(a+b)=18k-13k=5k$. Using the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(6k)^2+(5k)^2-(7k)^2}{2(6k)(5k)} = \frac{36k^2+25k^2-49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5}$. $\cos B = \frac{c^2+a^2-b^2}{2ca} = \frac{(5k)^2+(7k)^2-(6k)^2}{2(5k)(7k)} = \frac{25k^2+49k^2-36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35}$. $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{(7k)^2+(6k)^2-(5k)^2}{2(7k)(6k)} = \frac{49k^2+36k^2-25k^2}{84k^2} = \frac{60k^2}{84k^2} = \frac{5}{7}$. Therefore,$\cos A+\cos B+\cos C = \frac{1}{5}+\frac{19}{35}+\frac{5}{7} = \frac{7+19+25}{35} = \frac{51}{35}$.

In $\triangle ABC$,with usual notations,if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$,then the value of $\cos A+\cos B+\cos C$ is

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