Hello, Moon1975!

A satellite is in orbit 800 km above the surface of the earth.

The satellite is viewed at an angle of 40 degrees above the earth's surface.

Calculate the distance between you and the satellite.

So far I have:

Triangle $\displaystyle ABC:$

. . point $\displaystyle A$: you

. . point $\displaystyle B$: centre of the Earth

. . point $\displaystyle C$: satellite

Length $\displaystyle AB$ is the **radius** of Earth, approx. 6370 km

Length $\displaystyle CB$ is the **radius** of Earth + 800km (altitude of satellite)

$\displaystyle \angle BAC$ is 90 + 40 = 130 degrees

However this triangle doesn't allow me to find the distance $\displaystyle AC.$

Yes, it does. Code:

C
o
*|
* |
* |
* | 800
* |
* |
* *
* |
* |
A o 130d |
* | 6370
* |
6370 * |
* |
o
B

Law of Sines: .$\displaystyle \frac{\sin C}{AB} \:=\:\frac{\sin A}{BC}$

. . $\displaystyle \sin C \:=\:\frac{6370\sin130^o}{7170} \:=\:0.685351559$

. . Hence: .$\displaystyle \angle C \:\approx\:43.263^o$

Then: .$\displaystyle \angle B \;=\;180^o - 130^o - 43.263^o \;=\;6.737^o$

Law of Sines: .$\displaystyle \frac{AC}{\sin6.737^o} \:=\:\frac{71670}{\sin130^o}$

. . $\displaystyle AC \:=\:\frac{7170\sin6.737^o}{\sin130^o} \:=\:1098.014065$

Therefore: .$\displaystyle AC \;\approx\;1098.0\text{ km.}$