With usual notations, if in a triangle ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then cos A : cos B : cos C is :
Answer & explanation
Correct answer: option 1
If each ratio be k, then we have Let b + c = 11k, c + a = 12k, a + b = 13k
So that 2(a + b + c) = 36k and a = 7k, b = 6k, c = 25k
Now $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{36+25-49}{2.6.5}=\frac{1}{5}$
Similarly, $\cos B=\frac{19}{35}$ and $\cos C = 5/7$; cos A : cos B : cos C = $\frac{1}{5} : \frac{19}{35} : \frac{5}{7} = 7 :19 : 25$.