Example 15 As shown in Figure 2-19, take a point $M$ inside $\triangle A B C$ such that $\angle M B A=30^{\circ}, \angle M A B=10^{\circ}$. Let $\angle A C B=80^{\circ}, A C=B C$, find $\angle A M C$.
(1983 Former Yugoslavia Olympiad Problem)
Let $\angle A C M=\alpha$, then $\angle M C B=80^{\circ}-\alpha$. By the trigonometric form of Ceva's Theorem, we have $\frac{\sin \alpha}{\sin \left(80^{\circ}-\alpha\right)} \cdot \frac{\sin 10^{\circ}}{\sin 40^{\circ}} \cdot \frac{\sin 20^{\circ}}{\sin 30^{\circ}}=1$.
Thus, $\sin \alpha \cdot \sin 10^{\circ}=\sin \left(80^{\circ}-\alpha\right) \cdot \cos 20^{\circ}$.
$$
\begin{array}{l}
\therefore 2 \sin \alpha \cdot \cos 80^{\circ}=2 \sin \left(80^{\circ}-\alpha\right) \cdot \cos 20^{\circ}, \\
\therefore \quad \sin \left(\alpha+80^{\circ}\right)+\sin \left(\alpha-80^{\circ}\right)=\sin \left(100^{\circ}-\alpha\right)+\sin \left(60^{\circ}-\alpha\right), \\
\therefore \sin \left(\alpha-80^{\circ}\right)-\sin \left(60^{\circ}-\alpha\right)=\sin \left(100^{\circ}-\alpha\right)-\sin \left(\alpha+80^{\circ}\right) \\
=2 \cos 90^{\circ} \cdot \sin \left(10^{\circ}-\alpha\right)=0. \\
\end{array}
$$

Thus, $\sin \left(\alpha-80^{\circ}\right)=\sin \left(60^{\circ}-\alpha\right)$.
Noting that $0<\alpha<80^{\circ}$, we have $-80^{\circ}<\alpha-80^{\circ}<0$ and $60^{\circ}-\alpha<60^{\circ}$.
$\therefore \quad \alpha-80^{\circ}=60^{\circ}-\alpha$, hence $\alpha=70^{\circ}$.
Therefore, $\angle A M C=180^{\circ}-\angle M A C-\angle A C M=180^{\circ}-40^{\circ}-70^{\circ}=70^{\circ}$ is the solution.
Note: This result can also be directly obtained from (1) as follows:
$$
\left\{\begin{array}{l}
\sin \alpha=\sin 70^{\circ} \\
\sin 10^{\circ}=\sin \left(80^{\circ}-\alpha\right)
\end{array} \text { and } 0<\alpha, 80^{\circ}-\alpha<80^{\circ} \text {, yielding } \alpha=70^{\circ}\right. \text {. }
$$

Alternatively, this problem can be solved as follows: From $\frac{\sin \alpha}{\sin \left(80^{\circ}-\alpha\right)} \cdot \frac{\sin 10^{\circ}}{\sin 40^{\circ}} \cdot \frac{\sin 20^{\circ}}{\sin 30^{\circ}}=1$, we have
$$
\frac{\sin \left(80^{\circ}-\alpha\right)}{\sin \alpha}=\frac{\sin 10^{\circ} \cdot \sin 20^{\circ}}{\sin 40^{\circ} \cdot \sin 30^{\circ}}=\frac{\sin 10^{\circ}}{\cos 20^{\circ}}=\frac{\sin \left(80^{\circ}-70^{\circ}\right)}{\sin 70^{\circ}}=\sin 80^{\circ} \cdot \cot 70^{\circ}-\cos 80^{\circ}.
$$

Since $\frac{\sin \left(80^{\circ}-\alpha\right)}{\sin \alpha}=\sin 80^{\circ} \cdot \cot \alpha-\cos 80^{\circ}$ is a strictly decreasing function of $\alpha$ in $\left(0^{\circ}, 180^{\circ}\right)$, we have $\angle A C M=\alpha=70^{\circ}$. Therefore, $\angle A M C=180^{\circ}-40^{\circ}-70^{\circ}=70^{\circ}$.

Or, from $\angle A M B=140^{\circ}$, let $\angle A M C=x$, then $\angle C M B=220^{\circ}-x$. Applying the trigonometric form of Ceva's Theorem to $\triangle M A B$ and point $C$, we have
$1=\frac{\sin \angle A M C}{\sin \angle C M B} \cdot \frac{\sin \angle M B C}{\sin \angle C B A} \cdot \frac{\sin \angle B A C}{\sin \angle C A M}=\frac{\sin x}{\sin \left(220^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 50^{\circ}} \cdot \frac{\sin 50^{\circ}}{\sin 40^{\circ}}$.
Thus, $\frac{\sin \left(220^{\circ}-x\right)}{\sin x}=\frac{1}{2 \cos 20^{\circ}}=\frac{\sin \left(220^{\circ}-70^{\circ}\right)}{\sin 70^{\circ}}=\sin 220^{\circ} \cdot \cot 70^{\circ}-\cos 220^{\circ}$.
Since $\frac{\sin \left(220^{\circ}-x\right)}{\sin x}=\sin 220^{\circ} \cdot \cot x-\cos 220^{\circ}\left(\sin 220^{\circ}<0\right)$ is a strictly increasing function of $x$ in $\left(0^{\circ}, 180^{\circ}\right)$,
we have $\angle A M C=x=70^{\circ}$.