For example, in $\triangle A B C$, $\angle A B C=60^{\circ}$, point $P$ is inside the triangle. Given $P A=10, P B=6, P C=7$. Find the maximum value of the area of $\triangle A B C$.
Solution: As shown in Figure 3, construct $\square ABCD$, and draw $PQ \perp BC$, connect $DQ, CQ, PD$. It is easy to see that
$$
\begin{array}{l}
DQ = PA = 10, \\
CQ = PB = 6.
\end{array}
$$

Let $PD = x$.
To find $\left(S_{\triangle ABC}\right)_{\text{max}}$, it is sufficient to find $\left(S_{\square ABCD}\right)_{\text{max}}$, but $\angle ABC = 60^{\circ}$, which means we need to find $(AB \cdot BC)_{\text{max}}$.

In the convex quadrilateral $PCQD$, applying the generalized Ptolemy's theorem, we have
$$
PQ \cdot CD \leq DQ \cdot PC + CQ \cdot PD,
$$

which means $AB \cdot BC \leq 70 + 6x$.
Obviously, when $P, C, Q, D$ are concyclic, i.e.,
$$
\angle PAB = \angle QDC = \angle QPC = \angle PCB,
$$

then
$$
AB \cdot BC = 70 + 6x.
$$

Connect $BD$ intersecting $AC$ at point $O$, and then connect $PO$.
In $\triangle PBD$ and $\triangle PAC$, $PO$ is their common median, thus,
$$
\begin{array}{l}
PB^2 + PD^2 = 2OB^2 + 2OP^2, \\
PA^2 + PC^2 = 2OA^2 + 2OP^2.
\end{array}
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
\left(PB^2 + PD^2\right) - \left(PA^2 + PC^2\right) \\
= 2\left(OB^2 - OA^2\right) \\
= \frac{1}{2}\left(BD^2 - AC^2\right) \\
= \frac{1}{2}\left(BD^2 + AC^2\right) - AC^2 \\
= AB^2 + BC^2 - AC^2 \\
= AB \cdot BC,
\end{array}
$$

which means
$$
\begin{array}{l}
AB \cdot BC = 6^2 + x^2 - 10^2 - 7^2 \\
= x^2 - 113.
\end{array}
$$

Comparing equations (1) and (2), we get
$$
x^2 - 6x - 183 = 0.
$$

Solving for $x$, we get $x = 3 + 8\sqrt{3}$ (negative value is discarded).
$$
\begin{array}{l}
\text{Therefore, }\left(S_{\triangle ABC}\right)_{\max} \\
= \frac{1}{2} AB \cdot BC \sin \angle ABC \\
= \frac{1}{2}(70 + 6x) \sin 60^{\circ} \\
= \frac{\sqrt{3}}{4}[70 + 6(3 + 8\sqrt{3})] \\
= 36 + 22\sqrt{3}.
\end{array}
$$

Incidentally, when $S_{\triangle ABC}$ reaches its maximum value, let $\angle PAB = \angle PCB = \alpha$, we can use the characteristics of the figure to find $\sin \alpha = \frac{3}{\sqrt{73}}$ and the lengths of $AB$ and $BC$ (by drawing $PM \perp AB$ at point $M$ and $PN \perp BC$ at point $N$, obviously, $P, M, B, N$ are concyclic, with the diameter of the circle being $PB = 6$, and $\left.\angle MPN = 120^{\circ}\right)$.