Question

With usual notations, if in a triangle, ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ then prove that $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$.

Answer 1

If each ratio be k, then we have
$b+c=11k,c+a=12k,a+b=13k$

so that,

$2(a+b+c)=36k$ or $a+b+c=18k$

$\therefore a=7k,b=6k,c=5k$

Now,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$=\frac{36+25-49}{\mathrm{2.6.5}}=\frac{12}{60}=\frac{1}{5}$

Similarlly, $\cos B=19/35$ and $\cos C=5/7$

$\therefore \cos A:\cos B:\cos C=\frac{1}{5}:\frac{19}{35}:\frac{5}{7}=7:19:25.$