point on line to form angle

Solution 1:

Given a line segment $\overline{AB}$, the locus of points $P$ such that $\angle{APB}$ is constant is a pair of circular arcs that pass through $A$ and $B$.

The centers of these arcs lie on the perpendicular bisector of $\overline{AB}$ at a distance of $\frac12|AB\cot\theta|$ from the line segment. This value comes from the fact that $\angle{ACB}$ is $2\theta$.

Once you have the two circles, it’s a straightforward computation to find their intersections with the camera rail. You’ll need to take a bit of care to reject solutions that are outside of these arcs: when $\theta$ is acute, you only want points on the circle that are on the same side of $\overline{AB}$ as the circle’s center—the solid arcs in the above illustration; when it’s obtuse, you want the arc on the opposite side—the broken arcs in the illustration. If $\theta=\pi/2$, then there’s effectively only one circle and its center is the midpoint of $\overline{AB}$. There can be as many as four solutions, two for each arc.