Example 9 As shown in Figure 12,
in $\triangle A B C$, $A C$ $=B C, \angle A C B=$ $80^{\circ}$, take a point $M$ inside $\triangle A B C$, such that $\angle M A B=10^{\circ}$, $\angle M B A=30^{\circ}$. Find the degree measure of $\angle A M C$.
(1983, Former Yugoslavia Mathematical Olympiad)
Solution 1: Since $A C=B C, \angle A C B=80^{\circ}$, then $\angle C A B=\angle C B A=50^{\circ}$.
Since $\angle M A B=10^{\circ}, \angle M B A=30^{\circ}$, therefore,
$$
\angle C A M=40^{\circ}, \angle C B M=20^{\circ} \text {. }
$$

Let $\angle B C M=x$, then, $\angle A C M=80^{\circ}-x$.
Applying the trigonometric form of Ceva's Theorem to $\triangle A B C$ and point $M$ we have
$$
\begin{array}{l}
1=\frac{\sin \angle C A M}{\sin \angle M A B} \cdot \frac{\sin \angle A B M}{\sin \angle M B C} \cdot \frac{\sin \angle B C M}{\sin \angle M C A} \\
=\frac{\sin 40^{\circ}}{\sin 10^{\circ}} \cdot \frac{\sin 30^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin x}{\sin \left(80^{\circ}-x\right)} . \\
\text { Then } \frac{\sin \left(80^{\circ}-x\right)}{\sin x}=\frac{\sin 40^{\circ} \cdot \sin 30^{\circ}}{\sin 10^{\circ} \cdot \sin 20^{\circ}} \\
=\frac{\cos 20^{\circ}}{\sin 10^{\circ}}=\frac{\sin \left(80^{\circ}-10^{\circ}\right)}{\sin 10^{\circ}} .
\end{array}
$$

Since $\frac{\sin \left(80^{\circ}-x\right)}{\sin x}=\sin 80^{\circ} \cdot \cot x-\cos 80^{\circ}$
is a strictly decreasing function of $x$ in $(0, \pi)$, therefore,
$$
\angle B C M=x=10^{\circ} \text {. }
$$

Thus, $\angle A M C=\angle M A B+\angle A B C+\angle B C M=70^{\circ}$.
Solution 2: Since $A C=B C, \angle A C B=80^{\circ}$, then
$$
\angle C A B=\angle C B A=50^{\circ} \text {. }
$$

Since $\angle M A B=10^{\circ}, \angle M B A=30^{\circ}$, therefore,
$$
\begin{array}{l}
\angle C A M=40^{\circ}, \angle C B M=20^{\circ}, \\
\angle A M B=140^{\circ} .
\end{array}
$$

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
$$
\begin{array}{l} 
1=\frac{\sin \angle A M C}{\sin \angle C M B} \cdot \frac{\sin \angle M B C}{\sin \angle C B A} \cdot \frac{\sin \angle B A C}{\sin \angle C A M} \\
=\frac{\sin x}{\sin \left(220^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 50^{\circ}} \cdot \frac{\sin 50^{\circ}}{\sin 40^{\circ}} . \\
\text { Then } \frac{\sin \left(220^{\circ}-x\right)}{\sin x}=\frac{1}{2 \cos 20^{\circ}} \\
=\frac{\sin \left(220^{\circ}-70^{\circ}\right)}{\sin 70^{\circ}} .
\end{array}
$$

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 $(0, \pi)$, therefore,
$$
\angle A M C=x=70^{\circ} .
$$