4. 196 On a horizontal plane, three points are 100 meters, 200 meters, and 300 meters away from the base of an antenna. The sum of the angles of elevation from these three points to the antenna is $90^{\circ}$. What is the height of the antenna?
[Solution] Let the height of the antenna be $x$ meters. The angles of elevation from points 100 meters, 200 meters, and 300 meters away from the base of the antenna are $\alpha, \beta, \gamma$ respectively. Then
$$\operatorname{tg} \alpha=\frac{x}{100}, \operatorname{tg} \beta=\frac{x}{200}, \operatorname{tg} \gamma=\frac{x}{300} .$$

Since $\alpha+\beta+\gamma=90^{\circ}$, we have
$$\begin{aligned}
\operatorname{tg} \gamma & =\frac{x}{300}=\operatorname{tg}\left[90^{\circ}-(\alpha+\beta)\right] \\
& =\operatorname{ctg}(\alpha+\beta)=\frac{1}{\operatorname{tg}(\alpha+\beta)} \\
& =\frac{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=\frac{1-\frac{x}{100} \cdot \frac{x}{200}}{\frac{x}{100}+\frac{x}{200}} \\
& =\frac{100 \times 200-x^{2}}{300 x} .
\end{aligned}$$

Therefore, $2 x^{2}=100 \times 200$, and since $x>0$, we have $x=100$ meters.