$10 \cdot 2$ On a plane, given $n(n \geqslant 3)$ points $A_{1}, A_{2}, \cdots, A_{n}$, where no three points are collinear. Let $\alpha$ denote the minimum value of all angles $\angle A_{i} A_{j} A_{k}$, where $A_{i}, A_{j}, A_{k}$ are three different given points. For each $n$, find the maximum value of $\alpha$, and determine when these points are distributed to achieve the maximum value.
[Solution] Consider the case where $n$ given points are the $n$ vertices of a regular $n$-sided polygon. In this case, each interior angle of the regular $n$-sided polygon is $\frac{n-2}{n} 180^{\circ}$. Connecting the diagonals $A_{1} A_{j}, j=3,4, \cdots, n-1$, then $\angle A_{2} A_{1} A_{n}$ is divided into $n-2$ equal angles. Thus, each angle is $\frac{180^{\circ}}{n}$, and it is easy to see that $\alpha=\frac{180^{\circ}}{n}$. Therefore, the maximum value of $\alpha$ is no less than $\frac{180^{\circ}}{n}$.

On the other hand, suppose the convex hull of the $n$ given points is an $m$-sided polygon, where $3 \leqslant m \leqslant n$. Then the sum of the interior angles of this convex $m$-sided polygon is $(m-2) 180^{\circ}$, and the smallest interior angle does not exceed $\frac{m-2}{m} 180^{\circ}$. Let this smallest interior angle be $\angle A_{1} A_{2} A_{3}$. Then the other $n-3$ given points are all inside this angle. Draw the rays $A_{2} A_{j}, j=4,5, \cdots, n$, then these rays divide $\angle A_{1} A_{2} A_{3}$ into $n-2$ angles, and the smallest angle among them does not exceed $\frac{(m-2)}{m(n-2)} 180^{\circ}$. Since $m \leqslant n$, we have $\frac{m-2}{m} \leqslant \frac{n-2}{n}$, thus $\frac{m-2}{m(n-2)} 180^{\circ} \leqslant \frac{180^{\circ}}{n}$, which means $\alpha \leqslant \frac{180^{\circ}}{n}$.

In summary, the maximum value of $\alpha$ is $\frac{180^{\circ}}{n}$, and it is achieved when the $n$ points are the $n$ vertices of a regular $n$-sided polygon.