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

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(A) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$. Then $b+c = 11k$,$c+a = 12k$,and $a+b = 13k$. Adding these equations,we get $2(a+b+c)

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$7 : 19 : 25$

Vedclass · Answer

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

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

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