4. In $\triangle A B C$, $\angle A B C=90^{\circ}, A C=\sqrt[3]{2}, D$ is a point on the extension of $A C$, $C D=A B=1$. Then the degree measure of $\angle C B D$ is $\qquad$
$4.30^{\circ}$.
As shown in Figure 4, draw $D E / / B C$ intersecting the extension of $A B$ at $E$, then $\angle C B D=\angle B D E$.
From $\frac{A C}{C D}=\frac{A B}{B E}$ we get
$$
\begin{array}{l}
B E=\frac{1}{\sqrt[3]{2}}=\frac{\sqrt{4}}{2} . \\
\begin{array}{l}
\left.\therefore D E^{2}=A i\right)-A E^{2} \\
=(\sqrt[3]{2}+1)^{2}-\left(\frac{\sqrt[3]{4}}{2}+1\right)^{2} \\
=\sqrt[3]{\frac{27}{4}} .
\end{array} \\
D E=\sqrt[3]{\frac{27}{4}} .
\end{array}
$$
$$
\therefore \operatorname{ctg} \angle C B D=\operatorname{ctg} \angle B D E=\frac{D E}{B E}=\frac{\sqrt{\frac{27}{4}}}{\frac{\sqrt[3]{4}}{2}}=\sqrt{3} \text {. }
$$

And $0^{\circ}<\angle C B D<90^{\circ}$,
$$
\therefore \angle C B D=30^{\circ} \text {. }
$$