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 $E$, 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 $A O$. Since $P A \perp A O, A E \perp P O$, we get $P A^{2}=P E \cdot P O$.
Also, $P A^{2}=P B \cdot P C$, so $P E \cdot P O=P B \cdot P C$.
Thus, $\frac{P E}{P C}=\frac{P B}{P O}$.
Since $\angle E P B=\angle C P O$, therefore,
$\triangle P E B \backsim \triangle P C O$.
Hence $\angle B E P=\angle B C O=30^{\circ}$.
It is easy to see that $\angle F E O=\angle B E P=30^{\circ}$.
In $\triangle E F O$,
$$
\begin{array}{l}
\angle E O F=180^{\circ}-\angle E F O-\angle F E O \\
=180^{\circ}-20^{\circ}-30^{\circ}=130^{\circ} .
\end{array}
$$

Therefore, the central angle corresponding to $\overparen{D C F}$ is
$$
360^{\circ}-\angle E O F=230^{\circ} \text {. }
$$

Thus, $\angle D A F=\frac{1}{2} \times 230^{\circ}=115^{\circ}$.