Indeed. Knowing a side ($\gamma$) and the two angles at either side ($A$ and $B$), we can use the dual Law of Cosines to get the third angle ($C$):
$$
\cos(C)=-\cos(A)\cos(B)+\sin(A)\sin(B)\cos(\gamma)
$$
Then, we can use the Law of Sines as usual to get the other sides.
Alternatively, we can use the same dual Law of Cosines to get
$$
\cos(\alpha)=\frac{\cos(A)+\cos(B)\cos(C)}{\sin(B)\sin(C)}
$$
and
$$
\cos(\beta)=\frac{\cos(B)+\cos(A)\cos(C)}{\sin(A)\sin(C)}
$$
Note that, unlike plane trigonometry, we can determine a spherical triangle by AAA.
On an $x,y$ Cartesian plane,
suppose we are given the coordinates of a point $C$, the radius $r$ of
a circle around $C$, the coordinates of a point $P$ outside the circle,
and a ray $\overrightarrow{PQ}$ starting at $P$ at an angle of $\theta$ clockwise from the $x$-axis. (The coordinates of $Q$ need not be known;
it is used here just as a way to name a particular ray starting at $P$.)
Find the distance $d$ from $P$ to $C$ and
the direction $\theta_0$ (clockwise from the
$x$-axis) of the ray $\overrightarrow{PC}$ from $P$ through $C$.
Compute the angle $\alpha$ between the rays $\overrightarrow{PQ}$ and $\overrightarrow{PC}$;
it is $\lvert \theta - \theta_0 \rvert$,
$\lvert \theta - \theta_0 + 2\pi \rvert$,
or $\lvert \theta - \theta_0 - 2\pi \rvert$,
whichever is smallest.
If $\alpha \geq \frac\pi2$ then there is no intersection, so we're done.
In all of the discussion below, assume that $\alpha < \frac\pi2$.
If $M$ is the point on ray $\overrightarrow{PQ}$ closest to $C$,
the triangle $PMC$ is a right triangle with a right angle at $M$
and the angle $\alpha$ at $P$.
Therefore the distance $CM = d \sin \alpha$
and the distance $PM = d \cos\alpha$.
(Note that we compute all of this without needing to find the coordinates of $M$.)
If $d \sin \alpha > r$ then there is no intersection.
If $d \sin \alpha = r$ then the ray $\overrightarrow{PQ}$ is tangent to the circle, $M$ is the point of tangency and is one endpoint of the line segment, and the distance $PM$ is $d \cos \alpha$.
If $d \sin \alpha < r$ then ray $\overrightarrow{PQ}$ intersects the circle in two points.
One of these points is between $P$ and $M$; call that point $X$.
Then $XMC$ is a right triangle with hypotenuse $XC = r$
and leg $CM = d \sin \alpha$; the length of the other leg is
therefore $XM = \sqrt{r^2 - d^2 \sin^2 \alpha}$.
Therefore
$$PX = PM - XM = d \cos \alpha - \sqrt{r^2 - d^2 \sin^2 \alpha},$$
which is the distance we were to find.
Best Answer
The density of points at an angle $\theta$ with respect to some fixed direction is $f(\theta)=\sin\theta$. Thus the cumulative distribution function is $F(\theta)=1-\cos\theta$, and thus the mean angle is
\begin{eqnarray} \int_0^\frac\pi2\left(1-F(\theta)\right)\mathrm d\theta=\int_0^\frac\pi2\cos\theta\,\mathrm d\theta=\left[\sin\theta\right]^\frac\pi2_0=1-0=1\;. \end{eqnarray}
In degrees, that’s $1\cdot\frac{180°}\pi\approx57.3°$, in agreement with your findings.