Example 5 Draw a semicircle with line segment $AB$ as the diameter, with the center of the circle being $O$. $C$ is a point on the semicircle. If $OC^2 = AC \cdot BC$, then $\angle CAB=$ $\qquad$
(1995, National Junior High School Mathematics League)
Solution: (1) As shown in the figure, draw a perpendicular line from point $C$ to $AB$, intersecting $AB$ at $D$. By the area formula and the given information, we have $CD = \frac{1}{2} OC$, thus $\angle COB = 30^{\circ}$.

Since $\triangle AOC$ is an isosceles right triangle, $\angle COD$ is an exterior angle of $\triangle AOC$, hence $\angle CAB = 15^{\circ}$.
(2) As shown in the figure, we know $\angle CBA$
$$
\begin{array}{l}
=15^{\circ}, \angle CAB=90^{\circ} \\
-\angle CBA=75^{\circ}.
\end{array}
$$

In summary, the answers are $15^{\circ}$ or $75^{\circ}$.