I have a bounty for this answer here
https://worldbuilding.stackexchange.com/questions/131610/what-would-be-used-for-coordinates-on-a-large-asteroid
The scenario: you are on an asteroid. Compasses do not work. You have 2 or more distant landmarks you can see. You have a map with all of these landmarks and the distances between each.
From your position, you cannot calculate a bearing relative to North (no compass) but you have a compass calipers and can calculate the angle to one landmark relative to another.
When I draw this I get one angle (your position and two distant landmarks) and the opposing side (known distance between the 2 landmarks). I don't think I can calculate the length of the other sides of the triangle. But if there are 3 more more landmarks, it looks to me like any given position within the triangle (or any other polygon made by the distant landmarks) is characterized by a series of triangles, each with an angle made up by my position and 2 of the distant landmarks.
Can I calculate my position with just the relative angles to distant landmarks of known position? No right triangles. No true north.
Best Answer
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $\frac \pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2\pi-2ADB$. If angle $ADB$ is less than $\frac \pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB \gt \frac \pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=\pi-ADB$ and $EF=\frac 12AB\tan (\pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.