In a triangle $A B C$, 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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Aicharya · Accepted Answer

$\begin{aligned} & \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k(	ext { say }) \ & \Rightarrow a+b+c=18 k \ & \Rightarrow a=7 k, b=6 k, c=5 k\end{aligned}$ Using cosine rule $\begin{aligned} & \cos A=\frac{1}{5}, \cos B=\frac{19}{35}, \cos C=\frac{5}{7} \ & \Rightarrow \cos A: \cos B: \cos C=\frac{1}{5}: \frac{19}{35}: \frac{5}{7}=7: 19: 25\end{aligned}$

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

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