Example 11 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=20^{\circ}$, $N$ is a point inside the triangle, $\angle N A B=40^{\circ}$, $\angle N B C=30^{\circ}$. Find the degree measure of $\angle N C B$.
Solution: As shown in Figure 11, draw a line through point $N$ perpendicular to the extension of $AC$ at $P$. On the extension of $AN$, take a point $Q$ such that $\angle QBC = 30^{\circ}$. Connect $C, Q, C, QB, PQ, PN$.

Since $\angle PAC = 70^{\circ} = \angle NAC$, point $P$ is the reflection of point $N$ over $AC$. Therefore,
$PC = NC$.

Given $\angle NAB = 40^{\circ}$ and $\angle ABC = 50^{\circ}$, we know $AQ \perp BC$. Thus, point $N$ is the reflection of point $Q$ over $BC$. Therefore, $QC = NC$.

Thus, $PC = QC$.

It is easy to see that $\triangle BQN$ is an equilateral triangle, so
$NB = NQ$.

Since $\angle NPA = 90^{\circ} - \angle PAC = 20^{\circ} = \angle NBA$, we have $NP = NB$.

Thus, $NP = NQ$.

It is easy to see that $\triangle PNC \cong \triangle QNC$. Therefore, $\angle NCP = \angle NCQ$, which means $2 \angle NCA = 2 \angle NCB$.

Thus, $\angle NCA = \angle NCB$.

Hence, $\angle NCB = \frac{1}{2} \angle ACB = 10^{\circ}$.