Example 2 In $\triangle A B C$, $\angle A B C=40^{\circ}, \angle A C B=20^{\circ}, N$ is a point inside $\triangle A B C$, $\angle N B C=30^{\circ}$, $\angle N A B=20^{\circ}$. Find the degree measure of $\angle N C B$.
Analysis: According to the Law of Sines, establish a trigonometric equation, and then solve the equation using trigonometric knowledge.

Solution: Let $\angle N C B=\alpha$, then $\angle N C A=20^{\circ}-\alpha$.

In $\triangle A B N$, $\angle A B N=\angle A B C-\angle N B C=10^{\circ}, \angle N A B=20^{\circ}$. By the Law of Sines, we have
$$
\frac{A N}{B N}=\frac{\sin 10^{\circ}}{\sin 20^{\circ}}
$$

In $\triangle B C N$, $\angle N B C=30^{\circ}, \angle N C B=\alpha$
By the Law of Sines, we have
$$
\frac{B N}{N C}=\frac{\sin \alpha}{\sin 30^{\circ}}
$$

In $\triangle A C N$,
$$
\angle N A C=100^{\circ}, \angle A C N=20^{\circ}-\alpha
$$

By the Law of Sines, we have
$$
\frac{N C}{A N}=\frac{\sin 100^{\circ}}{\sin \left(20^{\circ}-\alpha\right)}
$$

From (1), (2), and (3), we get
$$
\frac{\sin 10^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin \alpha}{\sin 30^{\circ}} \cdot \frac{\sin 100^{\circ}}{\sin \left(20^{\circ}-\alpha\right)}=1
$$

Therefore, $\sin \alpha=\sin \left(20^{\circ}-\alpha\right)$
Also, $0^{\circ}<\alpha<20^{\circ}$
Thus, $\alpha=20^{\circ}-\alpha$
Hence, $\alpha=10^{\circ}$, i.e., $\angle N C B=10^{\circ}$