4. As shown in Figure $2, P A$ is tangent to $\odot O$ at point $A, P C$ intersects $\odot O$ at points $B$ and $C, P O$ intersects $\odot O$ at point $D, A E \perp P O$ at point $E$. Connect $B E$ and extend it to intersect $\odot O$ at point $F$, connect $O C, O F, A D, A F$. If $\angle B C O=30^{\circ}, \angle B F O=20^{\circ}$, then the degree measure of $\angle D A F$ is $\qquad$.
4. $115^{\circ}$.

Connect $O A, C D, C F$.
Since $P A$ is tangent to $\odot O$ at point $A$, we have $O A \perp P A$.
By $A E \perp P O \Rightarrow P A^{2}=P E \cdot P O$.
Also, $P A^{2}=P B \cdot P C$, thus
$P E \cdot P O=P B \cdot P C$
$\Rightarrow C, B, E, O$ are concyclic
$\Rightarrow \angle F E O=\angle B C O=30^{\circ}$.
Also, $\angle B F O=20^{\circ}$, so
$$
\begin{array}{l}
\angle E O F=130^{\circ} \Rightarrow \angle F C D=65^{\circ} \\
\Rightarrow \angle D A F=180^{\circ}-\angle F C D=115^{\circ} .
\end{array}
$$