In a triangle ABC,with usual notations,if fra | vedclass

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(C) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$.Adding the equations: $2(a+b+c) = 36k \implies a+b+c = 18k$.Then $a = (a+b+c) - (b+c) = 18k -

Vedclass · Accepted Answer

$7:19:25$

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(C) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$.Adding the equations: $2(a+b+c) = 36k \implies a+b+c = 18k$.Then $a = (a+b+c) - (b+c) = 18k - 11k = 7k$.$b = (a+b+c) - (c+a) = 18k - 12k = 6k$.$c = (a+b+c) - (a+b) = 18k - 13k = 5k$.Using the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36k^2+25k^2-49k^2}{2(6k)(5k)} = \frac{12k^2}{60k^2} = \frac{1}{5}$.$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{49k^2+25k^2-36k^2}{2(7k)(5k)} = \frac{38k^2}{70k^2} = \frac{19}{35}$.$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{49k^2+36k^2-25k^2}{2(7k)(6k)} = \frac{60k^2}{84k^2} = \frac{5}{7}$.Thus,$\cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{5}{7} = \frac{7}{35} : \frac{19}{35} : \frac{25}{35} = 7 : 19 : 25$.

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

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