Here are some thoughts. Suppose that you have two points $A$ and $B$ which lie on a given circle of radius $r$. If $A$ and $B$ are diametrically opposed, then one can chose either half-circle (relative to the diameter $AB$) to get from $A$ to $B$, and the length of this path is $r\pi$. In general, with the aid of a picture, the arc length from $A$ to $B$ is $r\lvert\theta_B-\theta_A\rvert$, where $\theta_A$ and $\theta_B$ are the angles of $A$ and $B$, respectively. This is fairly easy to see by translating and scaling so your circle is centred at the origin and has radius $1$.
If one introduces a third point, $C$, then the total path length will be $r\lvert\theta_C-\theta_B\rvert + r\lvert\theta_B-\theta_A\rvert$, assuming, of course, that one traverses from $A$ to $B$, and then $B$ to $C$ (regardless of any change of direction at $B$.) If $A$, $B$ and $C$ satisfy $\theta_A<\theta_B<\theta_C$, then the arc length from $A$ to $B$ to $C$ is precisely $r(\theta_C-\theta_A)$. One must be careful if your arc "wraps over" the positive $x$-axis. (Adding another point at $(1,0)$ and the absolute angle above will help with that case.)
Note that your angles must be in radians for the above to work.
Edit: A bit more thinking about the problem yields the following. Rotate everything so that $A$ is at $(1,0)$ (so $\theta_A$ is now $0$, $\theta_B$ is now $\theta_B-\theta_A$ ($\pm2\pi$ if necessary) , and $\theta_C$ is now $\theta_C-\theta_A$ ($\pm2\pi$ if necessary)). Then $A$ and $C$ divide the circle into two arcs, say $C_1$ and $C_2$, and, unless $C$ is at $(-1,0)$, the arc lengths of $C_1$ and $C_2$ are different. Now $B$ lies on exactly one of these arcs, which we may assume is $C_1$.
- If $\theta_C>\pi$ and $\theta_B<\theta_C$ then $C_1$ is the longer of the two arcs.
- If $\theta_C>\pi$ and $\theta_B>\theta_C$ then $C_1$ is the shorter of the two arcs.
- If $\theta_C>\pi$ and $\theta_B>\theta_C$ then $C_1$ is the shorter of the two arcs.
- If $\theta_C>\pi$ and $\theta_B<\theta_C$ then $C_1$ is the longer of
the two arcs.
Now the geometry is done, we just need to figure out the length of the longer arc and the shorter arc. This is easy since the shorter arc is $r\theta_C$ if $0<\theta_C\leq \pi$, and $(2\pi - \theta_C)r$ if $\pi<\theta_C<2\pi$. (The other arc has length $2\pi r$ minus the shorter arc length.) This should be relatively simple to code.
Best Answer
Let the sectors be $\theta_1\pm\Delta\theta_1$ and $\theta_2\pm\Delta\theta_2$ in your notation. The angle between the sector centers is $\phi = |\mathrm{wrap}(\theta_1-\theta_2)|$, where $$\mathrm{wrap}(\theta) = \begin{cases} \theta & \text{if $-180^\circ\le\theta\le180^\circ$,} \\ \theta+360^\circ & \text{if $\theta < -180^\circ$,} \\ \theta-360^\circ & \text{if $\theta > 180^\circ$} \end{cases}$$ is the angle normalized to lie between $-180^\circ$ and $180^\circ$. Then the angle of overlap is simply $\Delta\theta_1+\Delta\theta_2-\phi$; if this is negative, there is no overlap.