We measured the elevation angles of a tower standing on a horizontal plane at $50 \mathrm{~m}$ and $100 \mathrm{~m}$ from the base of the tower. The sum of the measured angles is $45^{\circ}$. How tall is the tower?
Let the height of the tower be denoted by $x$, and the two angles of elevation by $\alpha$ and $\beta$ as shown in the figure. From the two right-angled triangles formed,

$$
\operatorname{tg} \alpha=\frac{x}{50}, \quad \operatorname{tg} \beta=\frac{x}{100},
$$

and $\alpha+\beta=45^{\circ}$.

It is known that

$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}
$$

Substituting the above values:

$$
1=\frac{\frac{x}{50}+\frac{x}{100}}{1-\frac{x}{50} \frac{x}{100}} .
$$

After rearranging, we obtain the following quadratic equation for $x$: $x^{2}+150 x-5000=0$, from which, considering only the positive roots, $x=28.07$; this is the height of the tower in meters.

![](https://cdn.mathpix.com/cropped/2024_05_02_e092d931528d9db81a71g-1.jpg?height=241&width=551&top_left_y=785&top_left_x=776)