7. Given $O$ is the circumcenter of $\triangle A B C$, $A D$ is the altitude on $B C$, $\angle C A B=66^{\circ}, \angle A B C=44^{\circ}$. Then, $\angle O A D=$ $\qquad$ .
7. As shown in Figure 3, we have
$$
\begin{array}{l}
\angle O A D=90^{\circ}-\angle A E F \\
=90^{\circ}-(\angle A B C+\angle C B F)=90^{\circ}-\angle C A D-\angle A B C \\
=\angle A C B-\angle A B C=180^{\circ}-2 \angle A B C-\angle C A B=26^{\circ}
\end{array}
$$