Example 7 In $\triangle A B C$, $\angle A B C=50^{\circ}, \angle A C B=20^{\circ}, N$ is a point inside $\triangle A B C$, $\angle N A B=40^{\circ}$. $\angle N B C=30^{\circ}$. Find the degree measure of $\angle N C B$.
Draw a perpendicular line from point $N$ to $AC$, intersecting the extension of $BA$ at $P$. Take a point $Q$ on the extension of $AN$ such that $\angle QBC = 30^{\circ}$. Connect $PC$, $QC$, $QB$, $AP$, and $PN$.

Since $\angle PAC = 70^{\circ} = \angle NAC$,

point $P$ is symmetric to point $N$ with respect to line $AC$.

Therefore, $PC = NC$.

Since $\angle ABC = 50^{\circ}$ and $\angle NAB = 40^{\circ}$,

$AQ \perp BC$.

Also, $\angle NBC = 30^{\circ} = \angle QBC$,

so point $Q$ is symmetric to point $N$ with respect to line $BC$.

Therefore, $QC = NC$.

Thus, $PC = QC$.

Since $\angle QBN = \angle BNQ = 60^{\circ}$,

$\triangle BQN$ is an equilateral triangle.

Therefore, $BN = NQ$.

Since $\angle NPA = 90^{\circ} - \angle PAC = 20^{\circ} = \angle NBA$,

$BN = NP$.

Thus, $NP = NQ$.

Therefore, $\triangle PNC \cong \triangle QNC$.

So, $\angle NCP = \angle NCQ$.

That is, $2 \angle NCA = 2 \angle NCB$.

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

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