In a triangle ABC , with usual notations, the sides $a, \mathrm{~b}, \mathrm{c}$ are such that they are…

Let $a$, $b$, and $c$ be the roots of $x^3-11x^2+38x-40=0$, which represent the sides of triangle $ABC$. Using Vieta's formulas:

$a+b+c=11$, $ab+bc+ca=38$, and $abc=40$.

Testing integer divisors of $40$ reveals $x=2$ is a root. Factoring gives $(x-2)(x-4)(x-5)=0$, so the sides are $2$, $4$, and $5$.

Evaluate $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ using the identity:
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}.$

Substitute values: $a^2+b^2+c^2=4+16+25=45$ and $abc=40$, yielding $\frac{45}{80}=\frac{9}{16}$.

The result is $\boxed{\frac{9}{16}}$.