$$
\begin{array}{l} 
\text { 3. In quadrilateral } A B C D, \angle A B D=70^{\circ}, \\
\angle C A D=20^{\circ}, \angle B A C=48^{\circ}, \angle C B D=40^{\circ} \text {. Then } \\
\angle A C D=
\end{array}
$$
3. $22^{\circ}$.

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

which means that in this circle, the arcs $\overparen{AEC}$ and $\overparen{AE}$ subtend complementary inscribed angles.
Thus, $AE = BE$.

Also, $\angle EAC = \angle CBD = 40^{\circ}$, $\angle DAE = \angle EAC - \angle CAD = 20^{\circ} = \angle CAD$, so $\triangle ADC \cong \triangle ADE$.
Therefore, $\angle ACD = \angle AEB = 22^{\circ}$.