How to solve the triangulation problem? I have $3$ sensors. I've built the following system of equations that match the data from sensors. I need to find $x$, $y$, $R_a$, $R_b$, $R_c$, $\alpha$, $\beta$ and $\gamma$. Can you please help me, or at least hint how can I solve this system.
\begin{align}A_x & = x+R_a\cos \alpha \\
A_y & = y+R_a\sin \alpha \\
B_x & = x+R_b\cos \beta \\
B_y & = y+R_b\sin \beta \\
C_x & = x+R_c\cos \gamma \\
C_y & = y+R_c\sin \gamma \\
R_b-R_a & = z_1 \\ 
R_c-R_a & = z_2
\end{align}
 A: Hint: start with $R_b=z_1+R_a$. Then by rearranging the first two equations, squaring and adding, we can obtain: $R_a^2=(A-1)^2(x^2+y^2)$. Similarly, squaring and adding third and fourth equation, we get $(R_a+z_1)^2=(B-1)^2(x^2+y^2)$.Thus, $$\left(\frac{R_a}{R_a+z_1}\right)^2=\left(\frac{A-1}{B-1}\right)^2$$
A: From a trigonometric identity, the values of $x$ and $y$ must satisfy these equations simultaneously:
$$x^2+y^2 = \frac{R_a^2}{(A-1)^2}$$
$$x^2+y^2 = \frac{R_b^2}{(B-1)^2}$$
$$x^2+y^2 = \frac{R_c^2}{(C-1)^2}.$$
The only way for three circles with the same center (in this case, the origin) to have any real intersection points is for the circles to be identical:  that is:
$$\frac{R_a^2}{(A-1)^2}=\frac{R_b^2}{(B-1)^2}=\frac{R_c^2}{(C-1)^2}.$$
From the second to last equation,
$$\left|\frac{B-1}{A-1}\right|R_a-R_a = z_1$$
$$\implies R_a = \frac{z_1}{\left|\frac{B-1}{A-1}\right|-1}.$$
Therefore,
$$R_b = z_1+\frac{z_1}{\left|\frac{B-1}{A-1}\right|-1},\quad R_c = z_2+\frac{z_1}{\left|\frac{B-1}{A-1}\right|-1}.$$
This should give you a good start.
As far as finding the angles $\alpha,\beta,\gamma$, I'm not sure at the moment.  Perhaps someone else can shed light on this.
