4. In quadrilateral $A B C D$, $A B=\sqrt{2}, B C=\sqrt{3}$, $C D=1, \angle A B C=75^{\circ}, \angle B C D=120^{\circ}$. Then $\angle C D A=(\quad)$.
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $75^{\circ}$
(D) $90^{\circ}$
4. C.

As shown in Figure 3, construct a right triangle $\triangle B C E$ inward with $B C$ as one side, such that $C E=$
1. Then
$$
\begin{array}{l}
\angle C B E=30^{\circ}, \\
\angle B E C=60^{\circ}, \\
B E=2 .
\end{array}
$$

In $\triangle A B E$,
$$
\begin{array}{l}
\angle A B E=75^{\circ}-30^{\circ}=45^{\circ}, A B=\sqrt{2} . \\
\text { Therefore, } \angle A E B=45^{\circ} . \\
\text { In } \triangle C D E, C D=C E=1, \\
\angle D C E=120^{\circ}-90^{\circ}=30^{\circ} \\
\Rightarrow \angle C E D=\angle C D E=75^{\circ} \\
\Rightarrow \angle D E C+\angle C E B+\angle B E A=180^{\circ} \\
\Rightarrow \text { point } E \text { lies on } A D \\
\Rightarrow \angle C D A=\angle E D C=75^{\circ} .
\end{array}
$$