If in a triangle ABC,frac{b + c}{11} = frac{c | vedclass

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(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $ABC$.$\Rightarrow b+c=11K, c+a=12K, a+b=13K$.Adding these equations: $2(a+b+c)

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(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $ABC$.$\Rightarrow b+c=11K, c+a=12K, a+b=13K$.Adding these equations: $2(a+b+c) = 36K \Rightarrow a+b+c = 18K$.Subtracting each equation from the sum:$a = (a+b+c) - (b+c) = 18K - 11K = 7K$.$b = (a+b+c) - (c+a) = 18K - 12K = 6K$.$c = (a+b+c) - (a+b) = 18K - 13K = 5K$.Using the cosine rule:$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting the values:$\cos A = \frac{(6K)^2 + (5K)^2 - (7K)^2}{2(6K)(5K)} = \frac{36K^2 + 25K^2 - 49K^2}{60K^2} = \frac{12K^2}{60K^2} = \frac{1}{5}$.

If in a triangle $ABC$,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$,then $\cos A$ is equal to

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