8. (10 points) An angle can be represented by two capital letters on its two sides and the letter of its vertex, such as $\angle A O B$ in Figure 1 (the symbol “ $\angle$ ” represents an angle) and can also be represented by $\angle O$ (when there is only one angle at the vertex) as shown in Figure 2. In triangle $A B C$, $\angle B A O=\angle C A O, \angle C B O=\angle A B O, \angle A C O=\angle B C O, \angle A O C=110^{\circ}$, then $\angle C B O=$ $\qquad$
【Analysis】From the problem, we know that, as shown in the figure, in triangle $A O C$, $\angle A O C=110^{\circ}$. Since the sum of the interior angles of a triangle is $180^{\circ}$, we have $\angle C A O+\angle A C O=180^{\circ}-110^{\circ}=70^{\circ}$. Given that $\angle B A O=\angle C A O, \angle A C O=\angle B C O$, it follows that $\angle B A O+\angle B C O=\angle C A O+\angle A C O=70^{\circ}$, thus $\angle B A C+\angle B C A=70^{\circ}+70^{\circ}=140^{\circ}$. Therefore, $\angle A B C=180^{\circ}-140^{\circ}=40^{\circ}$. Since $\angle C B O=\angle A B O$, we have $\angle C B O=40^{\circ} \div 2=20^{\circ}$. This allows us to solve the problem.

【Solution】In triangle $A O C$,
$$
\begin{array}{l}
\angle C A O+\angle A C O=180^{\circ}-\angle A O C \\
=180^{\circ}-110^{\circ} \\
=70^{\circ},
\end{array}
$$

Since $\angle B A O=\angle C A O, \angle A C O=\angle B C O$,
we have $\angle B A O+\angle B C O=\angle C A O+\angle A C O=70^{\circ}$,
thus $\angle B A C+\angle B C A=70^{\circ}+70^{\circ}=140^{\circ}$,
in triangle $A B C$, $\angle A B C=180^{\circ}-(\angle B A C+\angle B C A)$
$$
\begin{array}{l}
=180^{\circ}-140^{\circ} \\
=40^{\circ},
\end{array}
$$

Therefore, $\angle C B O=\angle A B O=40^{\circ} \div 2=20^{\circ}$.
Answer: The measure of $\angle C B O$ is $20^{\circ}$.
So the answer is: $20^{\circ}$.