In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is an altitude of $\triangle ABC$, then what is $\cos \angle ABM$?
[asy]

pair A,B,C,M;

B = (0,0);

A = (0,10);

C = (24,0);

M = foot(B,A,C);

draw(M--B--A--C--B);

label("$B$",B,SW);

label("$A$",A,N);

label("$C$",C,SE);

label("$M$",M,NE);

draw(rightanglemark(C,B,A,30));

draw(rightanglemark(A,M,B,30));

[/asy]

The Pythagorean Theorem gives us $AC = \sqrt{AB^2 + BC^2} = \sqrt{100+576} = \sqrt{676}=26$.

From right triangle $ABM$, we have $\angle ABM = 90^\circ - \angle BAM = 90^\circ - \angle BAC$.  But right triangle $ABC$ gives us $90^\circ -\angle BAC = \angle BCA$, so $\angle ABM = \angle BCA$, which means $\cos \angle ABM = \cos\angle BCA = \frac{BC}{AC} = \frac{24}{26} = \boxed{\frac{12}{13}}$.