Example 6 The main control panel of a chemical plant is 1 meter wide, the bottom edge of the panel is 2 meters above the ground, and the eyes of the on-duty personnel are 1.2 meters above the ground when sitting. Where should the on-duty personnel sit to see the panel most clearly? And find the maximum angle of view at this position.
Solution: As shown in Figure 1, the distance from the person's leg to the ground $A E=1.2$, the height of the clock face $B C=1$, and the distance from the bottom of the clock face to the ground $C F=2$. Draw $A D \parallel E F$ intersecting $B F$ at $D$.

Then $D F=1.2, C D=2-1.2=0.8$.

Let $\angle B A D=\alpha, \angle C A D=\beta$, then the viewing angle $\theta=\angle B A C=\angle B A D-\angle C A D=\alpha-\beta$. Let the horizontal distance from the person's eye to the clock face $A D=x$. Then
$\operatorname{tg} \alpha=\frac{1.8}{x}, \operatorname{tg} \beta=\frac{0.8}{x}$,
$\operatorname{tg} \theta=\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \operatorname{tg} \beta}$
$$
\begin{array}{l}
=\frac{\frac{1.8}{x}-\frac{0.8}{x}}{1+\frac{1.8}{x} \cdot \frac{0.8}{x}}=\frac{1}{x+\frac{1.44}{x}} \\
\leqslant \frac{1}{2 \sqrt{x \cdot \frac{1.44}{x}}}=\frac{1}{2.4}=\frac{5}{12} .
\end{array}
$$

When $x=\frac{1.44}{x} \Rightarrow x=1.2$,
$$
\theta_{\text {min }}=\operatorname{arctg} \frac{5}{12} \text {. }
$$

In summary, when the person is sitting at a horizontal distance of 1.2 meters from the clock face, the view is the clearest, and the maximum viewing angle is $\operatorname{arctg} \frac{5}{12}$.