The simplest way to determine that is the equinox.

On the equinox, the sun is directly over the equator.

If you take the equator, at mid day, and a point at 45 degrees north (or south) at mid day, you can form a right angle, isosceles triangle.

This is because for the equator, the angle between that point 45 degrees north and the sun is 90 degrees, as it is straight up, and at 45 degrees north, the observed angle of elevation of the sun is 45 degrees.

As it is isosceles the 2 sides which aren't the hypotenuse need to be the same length.

As the distance between 45 degrees north and the equator is ~3000 miles (closer to 5000 km), that means the sun must be ~ 3000 km high.

The same can be done for stars, like Polaris, where you use the north pole and 45 degrees north, get the same kind of triangle and get the same height.

Where it falls to pieces is when you start using other locations as well.

For example, if you use the north pole and the equator, then one remains a right angle (the one the object is above, so the equator for the sun, the north pole for polaris), while the other is roughly 0.5 degrees (which in the RE model is due to refraction).

Now you need to use some more complex math, using tan, where tan(theta)=h/d, where theta is the angle that isn't a right angle triangle, h is the height of the object (i.e. how high it is above the observer) and d is the distance along the ground. That can be rearanged to give h=d*tan(theta)

So that now gives h=6000 miles (more accurately 10008 km)*tan(0.5 degrees) which works out to be roughly 52 miles or 87 km.

So much closer, but still over the same point.

If you use 45 degrees north and the point it isn't above (so the pole for the sun and the equator for polaris), you get a non-right angle triangle, where one angle is 0.5 degrees (at the pole or equator) and one is 135 degrees (at 45 degrees north).

However, beside that there is a right angle isosceles triangle.

What this means is that it will be above a point h (miles or km) away (so in a direction opposite the point with the 0.5 degree angle) 45 degrees north.

So now tan(theta)=h/(d+h)

(d+h)tan(theta)=h

d*tan(theta)+h*tan(theta)=h

d*tan(theta)=h-h*tan(theta)

d*tan(theta)=h*(1-tan(theta))

So h=d*tan(theta)/(1-tan(theta))

d is now 3000 miles, so this gives 26 miles or 44 km, above a location quite close to 45 degrees north.

So by using the same 3 points in different ways you get 3 different heights and 2 different locations.

When you start going into 3D it gets even worse.

Still on the equinox, The sun can be at an angle of elevation of 45 degrees, due east in one location, and 90 degrees around Earth, it is at an angle of elevation of 45 degrees due west.

This matches the 3000 miles due to the geometry, but puts it in a location off the equator, even though at the same time it is observed directly above the equator.

You can get worse, the sun can be setting due west, (for all observers along a meridian, with it being roughly 0.5 degrees angle of elevation for everyone), while half way around the world, it is rising due east. This (the due east and due west part) puts it completely off Earth. If you allow an error of 0.5 degrees, and just consider the equator, you have 2 right angle triangles, both having a line from the north pole (or axis of Earth) off to the sun, then each has a line coming from the pole to the equator (with a distance of 10000 km in the FE model, 6381 km in the RE model if I recall correctly), and the angle at the equator is 89.5 degrees, allowing it to rise slightly north of due east).

This now gives tan(89.5)=d/10000km

Thus d~1.1 million km, completely off Earth.

If you lower the margin of error to 0.1 degrees, then you have tan of 89.9 degrees, putting it 5.7 million km off Earth.