Solar Elevation Angles -- Anomaly?

Question

"The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc." - Wikipedia

The meridian is when the sun is directly facing earth and is at exactly 90° overhead at some point along the longitudinal line at a given location.

My question is, if the sun is 93 million miles away and travels overhead at 90° between the Tropic of Capricorn and the Tropic of Cancer, how is it that there are Meridian solar elevation angles reported from relatively nearby cities on Earth of less than 89.99°?

Let's take for instance Calgary, Canada as of today May, 04 2016 -- estimated ground distance from Calgary to where the Sun is at 90° on Calgary's longitudinal line is ~3000 miles. The solar elevation angle (aka "altitude") is reported by the timeanddate website to be 55°! - TimeandDate

Reversing the angle, using right triangle math:

The base angle, our solar elevation angle, of a right angle triangle is:
    "solar elevation angle"= arctangent (h/a)
    89.998151749049 = arctangent (93000000/3000)

So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

That makes sense, since if something is 93 million miles away from us and you think of it as a triangle, then the distance between any place on earth and the location on earth where the sun is at exactly 90° overhead -- you should expect the other angle of the triangle to be at near 90°, connected by a relatively tiny sliver of distance in a much elongated triangle to the center point of the distant sun.

If we go the other direction and leave out the presumption that the sun is 93 million miles away. We take h as unknown, we know the reported solar elevation angle of 55°, we can again use right triangle math:

    Let a=3000, solar elevation angle=55°
    h = a * tangent ("solar elevation angle")
    4,284.4440202263 = 3000 * tangent(55°)

So according to my calculations, based on the advertised solar elevation angle of 55° at Calgary, Canada on May, 04 2016 at Meridian, at a ground distance of 3000 miles from where the Sun is at exactly 90° overhead --  the Sun is actually only 4,284.4440202263 miles high!

** Update **: The math would work out for a sphere Earth with a radius of 4000 miles, using the exact solar position and adding corrective angles due to a presumption of a spherical curvature:

    Calgary, Canada: 51°03′N 114°04′W
    Location of Sun at Calgary Meridian: 16°23′N 114°04′W
    Distance = 2395.386 miles
    Radius of Earth=4000 miles
    Arc distance = 4790.772 miles
    Width of arc = 4509.52059 miles
    Angle to center of the Earth=68.622754110849°
    68.622754110849°/2=34.31137705542450°

This math utilized calculations that may have had a bit less acurracy but to make it work you need to add a bandaid padding of approximately 34.31137705542450° to Calgary,Canada's advertised angles. If you do you arrive at 55°+34.31137705542450° = 89.31137705542450° which is close to what I would expect to see. If, we are truly living on a Flat Earth, then the calculations would be telling us that the Sun height is only 4,284.4440202263 miles high and we can use simple trignometry.
Arc Calculator
Coordinate Distance Calculator

If the math is difficult for you, you can check with an online triangle calculator such as: Ke!san

Answer 1

Your confusion may stem from the way we sometimes draw the Sun (not to scale) as though it's close to the Earth.

In reality, as you note, the lines of vision to the Sun are nearly parallel:

Note that the angles (measured from the red line, the horizon, to the arrow) are still different, but for entirely different reasons.

I've actually stolen @Andy's answer (and @Tanenthor's comment), but I'm hoping the diagram makes it easier to understand.

Re the parallel lines, the 11810x60 image below may help explain: even though the lines eventually converge (on an object about 10 times the distance to the moon, still nowhere near the distance to the Sun), they are nearly parallel when leaving Earth.

Answer 2

So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

This is the wrong way to do it. A point 3000 miles south will be pointing in a very different direction, because the earth is round and only about 8000 miles in diameter.

For example:

Earth is around 25,000 miles in circumference. A point 3000 miles away is 0.12 times the circumference away. (3000/25000.) The circumference is equivalent to 360 degrees (assuming the Earth is a perfect circle, which is almost true for its polar section.) So 0.12 times 360 degrees is a difference of forty-something degrees. This probably accounts for your difference (very roughly).

Additional comment about roughness of these calculations: in fact the Earth is not a perfect sphere. Drawing a great circle along the meridian, the Earth will not appear perfectly round because the equator bulges slightly (and the poles are flattened). It's only something like 30-40 miles difference, but bear in mind this has a slight effect when calculating angles by dividing the circumference, as I did above. (Still better than assuming a straight line though, which is what I think the question is doing.)