Intersecting great circles to find position Is it possible to find the intersection of two great circles when knowing the following:
A point $a$ on earth,
A point $b$ on earth, and
The bearings of $a$ and $b$ from an observer?
 A: Given two points in the plane, the locus of points from which those points subtend a given angle is comprised of two circular arcs which connect the two points and are symmetric across the line containing them.
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The vertical bar on the right divides $\pi$ radians into two supplements. From the points in the black arcs, the two points subtend the angle indicated by the black part of the bar, while from the points in the red arcs, the two points subtend the angle indicated by the red part of the bar. If we know which of the two points lies on the right and which on the left as well as the angle subtended, we know that we are on the dark black versus the light black arc.
We have a similar situation on the sphere.
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On the sphere, there are two dark black arcs which degenerate into a cross when the angle subtended by the two points is the same as the angle between the two points on the sphere. When the angle subtended by the two points is greater than the distance between the points, one dark black arc is between the two points and the other is between the their antipodes. However, when the angle subtended by the two points is less than the distance between the two points, the dark black arcs connect each of the two points with the antipode of the other.
Since we know the relative bearing of three points (the two points and a pole), we intersect three of these pairs of arcs. In all but exceptional arrangements, there will only be one intersection of all three, thus you can find your location.
Of course, there are still exceptional situations such as if one of the points is on the equator and the other is the north pole; any point on the equator sees these at bearing $0^\circ$ and $90^\circ$ or $0^\circ$ and $270^\circ$.
To generate the paths on the sphere above, I used the Law of Sines
$$
\sin(\beta)=\sin(B)\frac{\sin(\alpha)}{\sin(A)}
$$
and the following identity
$$
\tan\left(\frac{\gamma}{2}\right)=\tan\left(\frac{\alpha+\beta}{2}\right)\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}
$$

Further Results
I have been able to determine that the locus of points on the sphere from which two given points subtend a given angle is the intersection of an elliptic cylinder with the sphere. The axis of the cylinder is parallel to the vector sum of the two given points; the cylinder passes through the two given points; the major axis of the intersection of the cylinder and the equatorial plane contains the center of the sphere.
If the two given points are separated by an angle of $\alpha$ and subtend an angle of $\mathrm{A}$ from the locus, then for the cross-section of the ellipse:
$$
\begin{array}{rl}
\text{semi-major axis:}&\frac{\tan(\alpha/2)}{\sin(\mathrm{A})}\\[4pt]
\text{semi-minor axis:}&\frac{\sin(\alpha/2)}{\sin(\mathrm{A})}\\[4pt]
\text{center offset:}&\frac{\tan(\alpha/2)}{\tan(\mathrm{A})}
\end{array}
$$
Suppose $\alpha=\frac\pi3$ and $\mathrm{A}=\frac{3\pi}{4}$:
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