Ok, I've worked out a derivation that I can show people who are interested. Let's start with an image depicting the scenario:
This is 2d space (no time axis). The observer observes an object at the observed position which has since moved to the current position with velocity ##v##. The value ##d## indicates the object's current distance from the observer, while ##d_o## is the observed distance (the path the light from the observed position took to reach the observer). The angle ##\alpha## represents the angle between the relative velocity vector and the observer (or object, depending on your perspective). The angle ##\alpha_o## is the observed version of that angle by the observer. We note that the distance between the observed position and current position is velocity times the amount of time the light took to reach the observer, so it is equal to ##v d_o / c##. We can use ##\beta = \frac v c## here to end up with ##\beta d_o##.
We can solve for ##d_o## using the law of cosines. This gives us:
$$\begin{align} d^2_o & = d^2 + \beta^2d^2_o - 2d \beta d_o \cos \left(\pi-\alpha\right) \\
0 & = d^2_o - \beta^2d^2_o - 2d \beta d_o \cos \alpha - d^2 \\
0 & = d^2_o \left(1 - \beta^2 \right) - 2d \beta d_o \cos \alpha - d^2 \end{align}$$
(We used ##\cos \left(\pi - \alpha \right) = -\cos \alpha## in here). We now have a quadratic equation in ##d_o## and we can use the quadratic formula to solve it. This gives us:
$$d_o = \frac {2d \beta \cos \alpha \pm \sqrt {4 d^2 \beta^2 \cos^2 \alpha + 4d^2 \left(1-\beta^2 \right)}} {2 \left( 1-\beta^2 \right)}$$
We can cancel out the 2's (and 4's in the radical) easily, and also pull out ##d## from the whole equation. We also note that ##\frac 1 {1 - \beta^2} = \gamma^2## so we can use that as well. With that, we are at:
$$\begin{align} d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \cos^2 \alpha - \beta^2 + 1} \right) \\
d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \left( \cos^2 \alpha - 1 \right) + 1} \right) \\
d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \sin^2 \alpha + 1} \right) \end{align}$$
We note that the value in the radical will always be greater than or equal to 1, and the value of ##\beta \cos \alpha## will always be less than 1, so only the positive square root gives us a sensible answer (otherwise we have a negative distance), so we use the positive only. And this gets us to my original equation: $$\frac {d_o} d = \gamma^2 \left(\beta \cos \alpha + \sqrt{1 + \beta^2 \sin^2 \alpha} \right)$$
To get ##\alpha_o## we can use the law of sines:
$$\begin{align} \frac {\sin \left( \alpha - \alpha_o \right)} {\beta d_o} & = \frac {\sin \left( \pi - \alpha \right)} {d_o} \\
\sin \left( \alpha - \alpha_o \right) & = \beta \sin \alpha \\
\alpha - \alpha_o & = \arcsin \left( \beta \sin \alpha \right) \\
\alpha_o & = \alpha - \arcsin \left( \beta \sin \alpha \right) \end{align}$$
Which is the observed angle equation I posted as well.
Thank you for your attention, and I hope some find this interesting.