Example 8 The width of the football field is $a$ meters, and the width of the goal is $b$ meters. In a football match, a winger of Team A dribbles past an opponent along the touchline and advances straight forward. Try to find out how far from Team B's goal line the winger should shoot to maximize the angle of the shot, and calculate the maximum angle.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
As shown in Figure 4-1-2, let $O$ be the intersection of the sideline and the opponent's baseline, $A$ and $B$ be the two goalposts, and $O_{1}$ be the midpoint of $A B$. According to the problem, we have
$$
|O A|=\frac{1}{2}(a-b),|O B|=\frac{1}{2}(a+b),
$$

Let $C$ be the point from which the shot is taken, $\angle A C B=\alpha,|\triangle C|=x$, then
$$
\begin{array}{c}
\tan \angle O C B=\frac{\frac{1}{2}(a+b)}{x}, \\
\tan \angle O C A=\frac{\frac{1}{2}(a-b)}{x}, \alpha=\angle O C B-\angle O C A,
\end{array}
$$

Therefore, $\tan \alpha=\tan (\angle O C B-\angle O C A)$
$$
\begin{array}{l}
=\frac{\tan \angle O C B-\tan \angle O C A}{1+\tan \angle O C B \cdot \tan \angle O C A}=\frac{\frac{\frac{1}{2}(a+b)}{x}-\frac{\frac{1}{2}(a-b)}{x}}{1+\frac{\frac{1}{2}(a+b)}{x} \cdot \frac{\frac{1}{2}(a-b)}{x}} \\
=\frac{b}{x+\frac{\frac{1}{4}\left(a^{2}-b^{2}\right)}{x}} \leqslant \frac{b}{\sqrt{a^{2}-b^{2}}},
\end{array}
$$

Equality holds when $x=\frac{\frac{1}{4}\left(a^{2}-b^{2}\right)}{x}$, i.e., $x=\frac{1}{2} \sqrt{a^{2}-b^{2}}$. Therefore, the winger should take the shot from a distance of $\frac{1}{2} \sqrt{a^{2}-b^{2}}$ meters from the opponent's baseline to maximize the angle of the shot, and the maximum angle is
$$
\arctan \frac{b}{\sqrt{a^{2}-b^{2}}}.
$$