Example 6. $\triangle A B C$ is an isosceles triangle, with vertex angle $A=$ $80^{\circ}, O$ is a point inside $\triangle A B C$, $\angle O B C=10^{\circ}$, $\angle O C B=30^{\circ}$. Find $\angle O A B$. (1992, Huanggang Region, Hubei Province Junior High School Mathematics Competition)
Let $N$ be the intersection of the angle bisector of $\angle A$ and the extension of $CO$ (as shown in the figure), then $BN$
$$
\begin{array}{l}
=CN. \\
\therefore \angle BNC=180^{\circ}-2 \angle NCB=120^{\circ}. \\
\because \angle ACB=\frac{1}{2}\left(180^{\circ}-\angle A\right)=50^{\circ}, \\
\therefore \angle ABN=50^{\circ}-30^{\circ}=20^{\circ}. \\
\text { Also } \angle OBN=30^{\circ}-10^{\circ}=20^{\circ}, \\
\angle BNA=180^{\circ}-20^{\circ}-\frac{1}{2} \times 80^{\circ} \\
=120^{\circ},
\end{array}
$$
$\therefore \triangle ABN \cong \triangle OBN$. Hence $AB=OB$.
$$
\therefore \angle BAO=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ}.
$$