Ok, I'll be basing this on the model I've seen quoted most often where the sun rotates around the north pole and is 3000 miles above the surface of the earth.

I'll go with Berlin, it's at approx 52.5

^{o} north. It's 2590 miles south of the north pole.

At the summer solstice sunrise is at 03:43 and sunset is at 20:33, that means that a viewer in Berlin can observe the sun for 16 hours and 50 minutes of it's 24 hour orbit around the north pole, put another way it is visible for 252

^{o} of it's 360

^{o} orbit.

I forgot to label the circle as the suns path.

If we just concentrate on sun rise for now we can draw 2 triangles as follows

We know the hypotenuse and the internal angles for the top triangle so we know how long the other sides are. As we now know the length of 2 of the 3 sides of the larger triangle we can work out the internal angles etc.

So in FET the sun is 35

^{o} from north and 25

^{o} from the horizon when it can be seen rising above the horizon. I tried to draw this but it didn't come out very well.

Now if you're actually at 52.5

^{o} north then the sun rises 47.72

^{0} and sets at 312.28

^{0} which is pretty close by FE standards requiring only a 12

^{o} shift horizontally and a 25

^{o} vertically. This leads me to think that who ever came up with this model lived just a few degrees further south.