Given frac{b + c}{11} = frac{c + a}{12} = fra | vedclass

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(A) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$. Summing these gives $2(a+b+c) = 36k$,so $a+b+c = 18k$. Subtracting each original equat

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$(7, 19, 25)$

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(A) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$. Summing these gives $2(a+b+c) = 36k$,so $a+b+c = 18k$. Subtracting each original equation from this 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 $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36k^2+25k^2-49k^2}{2(6k)(5k)} = \frac{12k^2}{60k^2} = \frac{1}{5}$. Similarly,$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{49k^2+25k^2-36k^2}{2(7k)(5k)} = \frac{38k^2}{70k^2} = \frac{19}{35}$. And $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{49k^2+36k^2-25k^2}{2(7k)(6k)} = \frac{60k^2}{84k^2} = \frac{5}{7}$. Given $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma} = \lambda$,we have $\alpha = \frac{\cos A}{\lambda} = \frac{1}{5\lambda}$,$\beta = \frac{\cos B}{\lambda} = \frac{19}{35\lambda}$,$\gamma = \frac{\cos C}{\lambda} = \frac{5}{7\lambda}$. Multiplying by $35\lambda$,we get $\alpha : \beta : \gamma = 7 : 19 : 25$.

Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$,then the ordered triple $(\alpha, \beta, \gamma)$ is proportional to

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