In $\Delta ABC$, if $b+c:c+a:a+b=11:12:13$ then $\cos A:\cos B:\cos C=$

$b+c:c+a:a+b=11:12:13$

Let  $b+c=11x;a+c=12x;a+b=13x$

       $\begin{array}{l}a=7x\\ b=6x\\ c=5x\end{array}$

$\cos A:\cos B:\cos C=\frac{b^2+c^2-a^2}{2bc}:\frac{a^2+c^2{b}^2}{2ac}=\frac{a^2+c^2-b^2}{2bc}$
$=\frac{{\left(6x\right)}^2+{\left(5x\right)}^2-{\left(7x\right)}^2}{2\left(6x\right)\left(5x\right)}:\frac{{\left(7x\right)}^2+{\left(5x\right)}^2-{\left(6x\right)}^2}{2\left(7x\right)\left(5x\right)}:\frac{{\left(7x\right)}^2+{\left(6x\right)}^2-\left(5x\right)}{2\left(7x\right)\left(6x\right)}$

$=\frac{36x^2+25x^2-49x^2}{60x^3}:\frac{49x^2+25x^2-36x^2}{70x^2}:\frac{49x^2+36x^2-25x^2}{84x^2}$

$=\frac{12}{60}:\frac{38}{70}:\frac{60}{84}=21:57:75=7:19:25$