3. (10 points) As shown in the figure, given that $A B C D E F G H I$ is a regular nonagon, then $\angle D I G=$ $\qquad$ degrees.
【Analysis】We can use the sum of the interior angles of a nonagon and the sum of the interior angles of a triangle, draw auxiliary lines, and connect the center of the regular nonagon, then $O I=O D=O G$, which allows us to find the degree of $\angle D I G$.

【Solution】Solution: According to the analysis, as shown in the figure, $O$ is the center of the regular nonagon, then $O I=O D=O G$,
$$
\begin{array}{l}
\angle D I G=\angle D I O+\angle O I G \\
=\frac{1}{2}(\angle \mathrm{DIO}+\angle \mathrm{ODI}+\angle 0 \mathrm{IG}+\angle \mathrm{OGI} \\
=\frac{1}{2} \angle D O G \\
=\frac{1}{2} \times\left(360^{\circ} \div 9 \times 3\right) \\
=60^{\circ}
\end{array}
$$