jmc

Since $△ABC\sim △CBD$, we have $\frac{BC}{AB}=\frac{29^3}{BC}⟹B{C}^2=29^{3}AB$. It follows that $29^2|BC$ and $29|AB$, so $BC$ and $AB$ are in the form $29^{2}x$ and $29x^2$, respectively, where x is an integer. By the Pythagorean Theorem, we find that $A{C}^2+B{C}^2=A{B}^2⟹(29^{2}x{)}^2+A{C}^2=(29x^2{)}^2$, so $29x|AC$. Letting $y=AC/29x$, we obtain after dividing through by $(29x{)}^2$, $29^2=x^2-y^2=(x-y)(x+y)$. As $x,y\in \mathbb{Z}$, the pairs of factors of $29^2$ are $(1,29^2)(29,29)$; clearly $y=\frac{AC}{29x}\ne 0$, so $x-y=1,x+y=29^2$. Then, $x=\frac{1+29^2}{2}=421$. Thus, $\cos B=\frac{BC}{AB}=\frac{29^{2}x}{29x^2}=\frac{29}{421}$, and $m+n=\boxed{450}$.