Finding a point from 3 known landmarks and relative headings to each I am trying to solve the problem whereby a scanning device must determine its location by visually detecting three beacons with known locations in its environment.  The beacons and scanner are all in the same plane.  However, the scanner does not know its true orientation -- it is only able to determine the relative angles between each beacon.  In the image below, the beacons are points A, B, and C, each at known x,y coordinates.  The scanner is located at point S.  The known relative angles between beacons are $\Theta _{ab}$, $\Theta _{bc}$, and $\Theta _{ca}$.  S is not necessarily within the area bounded by points A, B, and C.
scanner and beacon geometry
I've tried solving this using vector dot products of the three vectors AS, BS, and CS to create three formulas and then solve them algebraically as follows.
$$AS \bullet BS = |AS||BS| cos \Theta _{ab} = (A_x-S_x)(B_x-S_x) + (A_y-S_y)(B_y-S_y)$$
$$BS \bullet CS = |BS||CS| cos \Theta _{bc} = (B_x-S_x)(C_x-S_x) + (B_y-S_y)(C_y-S_y)$$
$$CS \bullet AS = |CS||AS| cos \Theta _{ca} = (C_x-S_x)(A_x-S_x) + (C_y-S_y)(A_y-S_y)$$
Expanding the formulas for vector magnitude
$$ \sqrt {(A_x-S_x)^2+(A_y-S_y)^2} \times \sqrt {(B_x-S_x)^2 + (B_y-S_y)^2} \times cos \Theta _{ab} = (A_x-S_x)(B_x-S_x) + (A_y-S_y)(B_y-S_y)$$
$$ \sqrt {(B_x-S_x)^2+(B_y-S_y)^2} \times \sqrt {(C_x-S_x)^2 + (C_y-S_y)^2} \times cos \Theta _{bc} = (B_x-S_x)(C_x-S_x) + (B_y-S_y)(C_y-S_y)$$
$$ \sqrt {(C_x-S_x)^2+(C_y-S_y)^2} \times \sqrt {(A_x-S_x)^2 + (A_y-S_y)^2} \times cos \Theta _{ca} = (C_x-S_x)(A_x-S_x) + (C_y-S_y)(A_y-S_y)$$
From this point on, I tried to solve the three formulas algebraically, but this quickly turned into a complex fourth order polynomial.
I'm thinking there must be an easier way to solve this problem.  Also, I believe the solution could be extended to three dimensions where the known beacon positions are at some elevation above the plane of the scanner and the angle above the plane is measurable, but at this point I will settle for a 2D solution.
 A: A geometric solution comes from the fact that an angle inscribed in a circle
cuts off an arc whose angle is exactly twice the inscribed angle.
Construct an isoceles triangle on $AB$ opposite point $S$,
with angles of $\theta_{AB} - \frac\pi2$ at $A$ and $B$, and
hence angle $2(\pi - \theta_{AB})$ at the third vertex, $P$.
Further, construct a circle with center $P$ through $A$ and $B$.
Then $S$ lies on that circle.
Similarly, using the angle $\theta_{BC}$,
you can construct a circle that passes through $B$ and $C$,
and $S$ will lie on that circle too.
The circles intersect at two points; one is $B$ and the other is $S$.
The third angle gives you a way to construct a third circle to confirm
the location of $S$; but more importantly, it gives you a way to detect
whether $S$ is inside triangle $\triangle ABC$ or outside;
and if outside, on which side it lies
of each of the lines through $A$, $B$, and $C$.
You may need this information in order to place the centers of the circles
in the right places; knowing only the angle $\theta_{AB}$, for example,
there are two possible circular arcs on which $S$ might lie, each of
which is a mirror image of the other through the line $AB$.
A: "100 Great Problems in Elementary Mathematics," published by Dover Publications, is a translation of a German work, and is still in print.
Problem 40 [Annex to a Survey] discusses the Snellius-Pothenot problem, which is the very thing you need. In the discussion of the problem, five things are the "givens:" The distance from A to C; the distance from B to C; the angle BAC; the angle CSA; and the angle BSA. All of these are ħere given according to the labels you use in your diagram; when you consult the article, you will have to make the appropriate substitutions yourself. Your point labels A, B, C and S are, in the article, C, B, A and P. The distance between your A and your C is a in the article; the distance between your A and your B is b in the article. What you call $\theta_{CA}$, the article calls $\alpha$; what you call $\theta_{AB}$, the article calls $\beta$.
