5. In $\triangle A B C$, $\angle A B C=36^{\circ}, \angle A C B=42^{\circ}$, take a point $D$ on side $B C$ such that $B D$ is exactly equal to the radius of the circumcircle of $\triangle A B C$. Then $\angle D A C=$ $\qquad$ degrees.
5. $54^{\circ}$.

Let $O$ be the circumcenter of $\triangle ABC$. It is easy to see that $\angle BAC = 102^{\circ}$ is an obtuse angle. Therefore, $O$ is outside $\triangle ABC$. Connect $OA$ and let it intersect side $BC$ at $D_1$. We will now prove that $D_1$ coincides with $D$.
From Figure 5, we can see that,
$$
\begin{array}{l}
\angle AOB = 84^{\circ}, \\
\angle OAB = 48^{\circ}, \\
\angle BD_1O = 36^{\circ} + 48^{\circ} = 84^{\circ}.
\end{array}
$$

Thus, $BD_1 = BO$, which means $D_1$ coincides with $D$. Therefore,
$$
\angle DAC = 102^{\circ} - 48^{\circ} = 54^{\circ}.
$$