In a Delta ABC, if frac{b + c}{11} = frac{c + | vedclass

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(B) Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = \lambda.$Then $b + c = 11\lambda$ $(i), c + a = 12\lambda$ $(ii),$ and $a + b = 13\l

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$\frac{5}{7}$

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(B) Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = \lambda.$Then $b + c = 11\lambda$ $(i), c + a = 12\lambda$ $(ii),$ and $a + b = 13\lambda$ $(iii).$Adding $(i), (ii),$ and $(iii),$ we get $2(a + b + c) = 36\lambda,$ so $a + b + c = 18\lambda$ $(iv).$Subtracting $(i)$ from $(iv),$ $a = 7\lambda.$Subtracting $(ii)$ from $(iv),$ $b = 6\lambda.$Subtracting $(iii)$ from $(iv),$ $c = 5\lambda.$Using the cosine rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}.$Substituting the values,$\cos C = \frac{(7\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2(7\lambda)(6\lambda)} = \frac{49\lambda^2 + 36\lambda^2 - 25\lambda^2}{84\lambda^2} = \frac{60\lambda^2}{84\lambda^2} = \frac{5}{7}.$

In a $\Delta ABC,$ if $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13},$ then $\cos C = $

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