3. In quadrilateral $A B C D$, $\angle A B D=70^{\circ}$, $\angle C A D=20^{\circ}$, $\angle B A C=48^{\circ}$, $\angle C B D=40^{\circ}$. Then $\angle A C D=$ $\qquad$
3. $22^{\circ}$.

As shown in Figure 5, construct the circumcircle of $\triangle A B C$, and let it intersect the extension of $B D$ at point $E$. Connect $A E$.
It is easy to find that $\angle A B C=110^{\circ}$,
$$
\begin{array}{l}
\angle A E B=\angle A C B \\
=22^{\circ}, \\
\angle A B C+\angle A B D \\
=180^{\circ},
\end{array}
$$

which means that in this circle, the arcs $\overparen{A E C}$ and $\overparen{A E}$ subtend complementary inscribed angles.
Thus, $A E=B E$.
$$
\begin{array}{l}
\text { Also, } \angle E A C=\angle C B D=40^{\circ}, \\
\angle D A E=\angle E A C-\angle C A D=20^{\circ}=\angle C A D,
\end{array}
$$

Therefore, $\triangle A D C \cong \triangle A D E$.
$$
\text { Hence } \angle A C D=\angle A E B=22^{\circ} \text {. }
$$