8. As shown in Figure 6, in $\triangle A B C$, the median $C M$ and the altitude $C D$ trisect $\angle A C B$. Then the measure of $\angle B$ is $\qquad$
8. $30^{\circ}$.

According to the problem, we have
$$
\begin{array}{l}
C D \perp A B, A M=M B, \\
\angle A C D=\angle M C D=\angle B C M .
\end{array}
$$

Since $\angle A C D=\angle M C D, C D \perp A M$, therefore,
$$
A D=D M=\frac{1}{2} A M \text {. }
$$

As shown in Figure 13, draw $M N \perp B C$ at point $N$.
Since $\angle D C M$
$$
\begin{array}{c}
=\angle N C M, \\
\angle C D M \\
=\angle C N M=90^{\circ},
\end{array}
$$

thus, $D M=N M$.
Therefore, $N M=\frac{1}{2} M B$.
Hence, in the right triangle $\triangle M N B$, $\angle B=30^{\circ}$.