In acute $\triangle ABC$, $AB > AC$, $\angle ACB - \angle ABC = 30^{\circ}$, $O$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$ respectively, and $\triangle DOH$ is an equilateral triangle. Find the value of $\angle DAO + \angle AOH$.
Draw $OE \perp AB$ at point $E$. Then $\angle C=\angle AOE=90^{\circ}-\angle EAO$.
Given that $H$ is the orthocenter of $\triangle ABC$, we have
$$
\begin{array}{l}
\angle B=90^{\circ}-\angle BAH. \\
\text { Therefore, } \angle C-\angle B=\angle BAH-\angle EAO \\
=\angle OAH=30^{\circ} .
\end{array}
$$

Since $\triangle DOH$ is an equilateral triangle, we have
$$
\angle ODH=60^{\circ}, DO=DH \text {. }
$$

Also, $\angle ODH=2 \angle OAH=60^{\circ}$, so points $O, H, A$ lie on the circumcircle with $D$ as the center and $DO$ as the radius.
$$
\begin{array}{l}
\text { Therefore, } DA=DO \\
\Rightarrow \angle DAO+\angle AOH=\angle DOA+\angle AOH \\
=\angle DOH=60^{\circ} .
\end{array}
$$