Question 55, Given in $\triangle ABC$, $AB=2, \frac{2}{\sin A}+\frac{1}{\tan B}=2 \sqrt{3}$, find the minimum value of the area of $\triangle ABC$.


Question 55, Solution: As shown in the figure, let's assume that $C$ is above $AB$, and $B$ is to the right of $A$. Let the foot of the perpendicular from $C$ to $AB$ be $O$, and establish a Cartesian coordinate system with $O$ as the origin and $\overline{OC}$ as the positive direction of the $y$-axis. Let $A(a, 0), B(b, 0), C(0, c)$, then $b - a = 2, \sin A = \frac{c}{\sqrt{a^2 + c^2}}, \tan B = \frac{c}{b}$. At this point, $S_{\triangle ABC} = \frac{2 \times c}{2} = c$. According to the given conditions:
$$
\begin{array}{l}
2 \sqrt{3} = \frac{2}{\sin A} + \frac{1}{\tan B} = \frac{2 \sqrt{a^2 + c^2}}{c} + \frac{b}{c} = \frac{2 \sqrt{a^2 + c^2} + (2 + a)}{c} \\
\Rightarrow 2 \sqrt{3} c = 2 + \sqrt{(1 + 3)(a^2 + c^2)} + a \geq 2 + (|a| + \sqrt{3} c) + a \geq 2 + \sqrt{3} c \\
\Rightarrow c \geq \frac{2 \sqrt{3}}{3}
\end{array}
$$

Therefore, $S_{\triangle ABC} \geq \frac{2 \sqrt{3}}{3}$, with equality when $A\left(-\frac{2}{3}, 0\right), B\left(\frac{4}{3}, 0\right), C\left(\frac{2 \sqrt{3}}{3}\right)$.