The farthest distance a satellite signal can directly reach is the length of the segment tangent to the curve of Earth’s surface. The diagram is not drawn to scale. If the angle formed by the tangent satellite signals is 128 degrees, what is the measure of the intercepted arc on Earth?
We can start by drawing a diagram. Let's call the center of the Earth point $O$, and let the tangent segment be $AB$:
```
A
\
\
\
\
O ----------- B
```
We know that the angle formed by the tangent lines (the angle $AOB$) is 128 degrees. We want to find the length of the intercepted arc on Earth, which is the length of the red arc in the following diagram:
```
Arc
| ,-.
| / \
|/ \
O-------P
```
To find the length of this arc, we need to first find the radius of the circle. Let's call it $r$. We also notice that $\angle AOB$ is an inscribed angle, meaning it is half the measure of the intercepted arc (arc $APB$). So arc $APB$ has a measure of $2 \cdot 128 = 256$ degrees.
Now we can use the formula for the circumference of a circle, $C = 2\pi r$, to find the length of the entire circle. Since the entire circle has an arc of 360 degrees, we can set up a proportion:
$$\frac{256}{360} = \frac{\text{length of red arc}}{2 \pi r}$$
Solving for the length of the red arc, we get:
$$\text{length of red arc} = \frac{256 \cdot 2 \pi r}{360} = \frac{32 \pi r}{5}$$
So all we need to find now is the radius $r$ of the circle. To do that, we can use some trigonometry. We notice that triangle $OAB$ is an isosceles triangle, since both sides ($OA$ and $OB$) are radii of the circle. So the angles $OAB$ and $OBA$ are equal, and each one is half of $\angle AOB$, or 64 degrees. We can use the tangent function to find the length of $OA$ (which is also the length of $OB$, since the triangle is isosceles):
$$\tan 64 = \frac{OA}{AB/2}$$
Solving for $OA$, we get:
$$OA = \frac{AB}{2} \cdot \tan 64 = \frac{AB}{2} \cdot 2.086$$
Now we just need to find the length of $AB$, which we could get from the diagram by using the fact that it is tangent to the curve of the Earth's surface. However, since the diagram is not drawn to scale, we don't have that information. So we could make a guess based on typical satellite orbits, or we could leave the answer in terms of $AB$.
Putting it all together, we get:
$$\text{length of red arc} = \frac{32 \pi OA}{5} = \frac{16 \pi AB \cdot \tan 64}{5}$$
This is our answer, in terms of $AB$.