## 22. Lighthouse

From the tower of the lighthouse, located 125.7 m above sea level, the horizon line is visible. At what approximate distance from the tower is this line, if we assume that the Earth's surface is spherical and its great circle is 40,000 km?

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-057.jpg?height=700&width=1334&top_left_y=889&top_left_x=361)
22. Let $S$ be the top of the lighthouse, $A$ be the point on the horizon, and $O$ be the center of the Earth. Triangle $O A S$ is a right triangle because line $S A$ is tangent to the spherical surface of the Earth. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-163.jpg?height=452&width=583&top_left_y=1353&top_left_x=745)

$$
O S^{2}=A O^{2}+A S^{2}
$$

But (in meters)

$$
O A=\frac{40000000}{2 \pi}
$$[^27]a

$$
O S=O A+125.7 \approx \frac{40000000}{2 \pi}+40 \pi
$$

Therefore,

$$
A S=\sqrt{O S^{2}-O A^{2}} \approx \sqrt{2 \frac{40 \pi \cdot 40000000}{2 \pi}} \approx 40000
$$

Thus, the horizon is approximately 40 km from the lighthouse.