## Task B-4.4.

Three points are located at distances of 100, 200, and 300 meters from the base of a vertical tower and lie in the same plane together with the base of the tower. Determine the height of the tower if the sum of the angles at which its top is seen from these points is $90^{\circ}$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-27.jpg?height=531&width=346&top_left_y=548&top_left_x=775)

Sketch

Let $\alpha_{1}$ be the angle at which the top of the tower is seen from a point horizontally 100 meters away from the base. Then, $\operatorname{tg} \alpha_{1}=\frac{h}{100}$. Similarly, if $\alpha_{2}$ is the angle at which the top of the tower is seen from a point horizontally 200 meters away, and $\alpha_{3}$ is the angle at which the top of the tower is seen from a point horizontally 300 meters away, then $\operatorname{tg} \alpha_{2}=\frac{h}{200}$ and $\operatorname{tg} \alpha_{3}=\frac{h}{300}$.

Since $\alpha_{1}+\alpha_{2}+\alpha_{3}=90^{\circ}$, applying the addition formulas, we get:

$$
\operatorname{tg} \alpha_{3}=\operatorname{tg}\left(90^{\circ}-\left(\alpha_{1}+\alpha_{2}\right)\right)=\operatorname{ctg}\left(\alpha_{1}+\alpha_{2}\right)=\frac{1}{\operatorname{tg}\left(\alpha_{1}+\alpha_{2}\right)}=\frac{1-\operatorname{tg} \alpha_{1} \operatorname{tg} \alpha_{2}}{\operatorname{tg} \alpha_{1}+\operatorname{tg} \alpha_{2}}
$$

Substituting the previously mentioned expressions for the tangents, we get:

$$
\begin{aligned}
\frac{h}{300} & =\frac{1-\frac{h}{100} \cdot \frac{h}{200}}{\frac{h}{100}+\frac{h}{200}} \\
\frac{h}{300} & =\frac{20000-h^{2}}{300 h} \\
h^{2} & =10000
\end{aligned}
$$

Thus, the height of the tower is 100 meters.