We have a triangle $\triangle ABC$ where $AC = 17,$ $BC = 15,$ and $AB = 8.$ Let $M$ be the midpoint of $AB.$ What is the length of $CM$?
First, we notice that $8:15:17$ is a Pythagorean triple, so we have a right triangle. Let's draw a sketch: [asy]
pair A, B, C, M;
A = (0, 8);
B = (0, 0);
C = (15, 0);
M = 0.5 * A + 0.5 * B;
draw(A--B--C--cycle);
draw(C--M);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, NE);
label("$8$", A--B, W);
label("$15$", B--C, S);
label("$17$", A--C, NE);
draw(rightanglemark(A,B,C,20));
[/asy] Since we know that $BM = \frac{1}{2} \cdot AB = 4,$ we simply apply the Pythagorean theorem to right triangle $\triangle MBC.$ \begin{align*}
CM^2 &= BM^2 + BC^2\\
CM^2 &= 4^2 + 15^2 = 241\\
CM &= \boxed{\sqrt{241}}.
\end{align*}