3B. An astronaut who is at a height of $525 \mathrm{~km}$ is looking at a part of the Earth. If the radius of the Earth is approximately $6300 \mathrm{~km}$, calculate the radius of the circular contour of the part of the Earth visible to the astronaut?
Solution. The radius $x$ of the circular contour of the part of the Earth visible to the astronaut is the height $BD$ to the hypotenuse $OA$ in the right-angled triangle $ABO$, where $\overline{OB}=6300 \text{ km}$ and $\overline{OA}=6300+525=6825 \text{ km}$. Point $O$ is the center of the Earth, and the astronaut is located at point $A$ (see diagram). Then,

![](https://cdn.mathpix.com/cropped/2024_06_05_a5e9539585a8c87968b2g-03.jpg?height=409&width=531&top_left_y=937&top_left_x=951)

$$
\overline{AB}=\sqrt{\overline{OA}^2-\overline{OB}^2}=\sqrt{6825^2-6300^2}=2625 \text{ km}
$$

From the similarity of triangles $ABO$ and $ADB$, both being right-angled and having one common angle, we have $\overline{OA}: \overline{OB}=\overline{BA}: \overline{BD}$, that is,

$$
6825: 6300=2625: x,
$$

from which we get $x=\frac{44100}{91} \text{ km}$.