I will work this problem from a pure mathematical perspective for the most part (using an abstract sphere rather than the Sun and radians rather than degrees), but at the end we can relate it to your specific problem.
A key fact in this answer is that the points on a sphere that project onto the circumference of the sphere's image in an orthogonal projection are a great circle on the sphere, specifically, the great circle you get when you slice the sphere with a plane that passes through the sphere's center and is parallel to the projection plane.
So your question comes down to, if we know the latitude of a point on that great circle, what is the longitude of that point?
Let's start by defining coordinates $\theta$ and $\lambda$ on the surface of the sphere. For consistency with mathematical formulas derived elsewhere, we will define $\theta$ as the colatitude (the angle measured along the surface from the "north pole" of the sphere) and $\lambda$ as the longitude (the angle that the point's line of longitude makes from the line of zero longitude). Working with colatitude rather than latitude may seem a little unusual, but that's the way many spherical geometry formulas are written, and it's easy to convert between colatitude and latitude when we need to.
In these coordinates, one equation describing a great circle
(adapted from a post on MathOverflow) is
$$ \cot\theta = \tan\theta_0 \cos(\lambda - \lambda_0) $$
where $\theta_0$ and $\lambda_0$ are the colatitude and longitude of a pole of the great circle.
(A pole of a great circle is a point that has the same relation to the great circle as the north and south poles have to the equator.)
Now, since we actually want to work with latitude rather than colatitude, we define the latitude $\phi$ as the angular distance on the sphere from the equator, so
$\phi = \frac\pi2 - \theta$ and $\phi_0 = \frac\pi2 - \theta_0.$
With these substitutions, the equation of the great circle is
$$ \tan\phi = \cot\phi_0 \cos(\lambda - \lambda_0). \tag1$$
So, given the latitude $\phi$ of a line of latitude on the sphere,
we can solve Equation $(1)$ for $\lambda$:
\begin{align}
\cos(\lambda - \lambda_0) &= \tan\phi_0 \tan\phi, \\
\lambda - \lambda_0 &= 2\pi n \pm \arccos(\tan\phi_0 \tan\phi)
&& \text{where $n$ is an integer}, \\
\lambda &= 2\pi n + \lambda_0 \pm \arccos(\tan\phi_0 \tan\phi)
&& \text{where $n$ is an integer}. \\
\end{align}
The reason for the multiple solutions is that there are two longitudes where the line of latitude intersects the great circle, and the longitude can be measured in various ways.
You might like to choose $n$ in such a way that $-\pi \leq \lambda \leq \pi,$
as we usually measure longitudes from $180$ degrees west to $180$ degrees east.
You actually have two choices for how to measure $\phi_0$ and $\lambda_0,$ because every great circle has two poles; but let's use the pole that we want to make visible in the projection. That is, $\phi_0$ and $\lambda_0$ are the latitude and longitude of the point that is projected into the exact center of the circle onto which the sphere is projected.
Note that for a sphere with no axial tilt relative to the projection plane (so the equator projects to a straight line), $\phi_0 = 0$;
if you tilt the axis so that the north pole is visible in the projection,
then $\phi_0 > 0$; and if you tilt the axis so that the south pole is visible in the projection, then $\phi_0 < 0.$
In the image in the question, for example, $\phi_0 < 0.$
You might also want to take into account the fact that when the axis of the sphere is not parallel to the plane of projection, the projections of the poles will be inside the circle onto which the sphere is projected, and you will want to plot one of those poles (whichever one is "visible" to the observer) at the correct projected point.
At the point where that pole is projected, the projected images of $360$ degrees' worth of lines of longitude should all meet. That is, you want it to look something like this image from stackoverflow:
In order to avoid projecting parts of lines of latitude from the "non-visible" side of the sphere onto the projected image, you will want to cut off the image of every line of longitude at the same great circle where the lines of latitude are cut off.
That is, given the longitude $\lambda$ of a particular line of longitude, you want to solve for $\phi$ in Equation $(1)$ in order to find the latitude where the line of longitude has to be cut off:
$$ \phi = \arctan(\cot\phi_0 \cos(\lambda - \lambda_0)). $$
Best Answer
https://en.wikipedia.org/wiki/Spherical_cap
The above article tells us that formula for the surface area of a spherical cap is $A = 2{\pi}r^2(1-cos\theta)$ where $\theta$ is the angle between the pole and the cap's end. In our case we are dealing with latitude, so what we want to eventually get is $90^{\circ} - \theta$.
The surface area of a sphere is $4{\pi}r^2$. We are working with a hemisphere, so we take half this to get $2{\pi}r^2$, and we want to find the point (line) in the middle where the areas are equal, which means we want half of this area, or $A = {\pi}r^2$. We can now form the equation ${\pi}r^2 = 2{\pi}r^2(1-cos\theta)$. We can divide both sides by ${\pi}r^2$ to get $1 = 2(1-cos\theta)$ then we can get $1/2 = 1-cos\theta$ then $1/2 = cos\theta$ then $acos(1/2) = \theta$ then $\theta = 60^{\circ}$.
This means that the average (mean) latitude on a sphere is $90^{\circ} - 60^{\circ}$ or $30^{\circ}$, which answers the stated question in the title. Additionally, to answer my use case, the average sun angular altitude at noon would be the same as the equinox angular altitude at noon, which is $90^{\circ} - latitude$ or $90^{\circ} - 30^{\circ}$ or $60^{\circ}$.