Finding angles of a triangle given a relation between it's sides and angles. F̶i̶n̶d̶ ̶a̶l̶l̶ ̶a̶n̶g̶l̶e̶s̶ (edit: that might have been asking for too much, sorry for the wrong problem statement, I thought that by solving for all angles, finding x would be trivial but as it seems to be thanks to timon92 that might not be possible) Find the angle $x$ knowing only the length of the side |AB|, the constants $E_A$ and $E_B$ and the relation $$E_B  r_B^2 sin(\gamma) = E_A r_A^2sin(\delta)$$


I've tried to solve a system of equations containing the known relation the sine law, the three cosine laws for each of the sides along with the pythagorean identities.


I wasn't even sure if this had a solution but I've managed to reduce it to a system of 4th degree polynomials and I'm still not sure if this is doable. Any sort of help would be greatly appreciated. (Edit: the second to last bracket is wrong but that doesn't change much, it's still a not an easy system of equations)
The give a bit of context the problem comes from my own project, where there is a light point source at C, two luminosity readings at A and B, $E_i = \frac{I}{r_i^2}cos(\theta) $, where I is the unknown intensity and $\theta$ is the angle between the illuminated surface and the plane perpendicular to the incident light. Originally I wanted to find the azimuth and altitude angles (x in the 2d case) of the light source given three sensors and their known relative position without the distance (so 2 out of 3 spherical coordinates) but even the 2d case stumped me.
 A: The problem is underconstrained. There are several triangles satisfying this condition.
By the sine law we have $\dfrac{r_A}{\sin \delta} = \dfrac{r_B}{\sin \gamma}$ , hence the condition is equivalent to
$$\frac{r_A}{r_B} = \left(\frac{E_B}{E_A}\right)^{1/3}.$$
The locus of points $C$ satisfying this equation is a certain circle (or a line if $E_A=E_B$) called Apollonius circle.
A: HINTS:
The polar relation
$$( r/a)^2= \sin \theta , \text{  which wlog can be written as :} $$
$$( r_A= \sqrt{\sin \gamma \;E_B},\; r_B= \sqrt{\sin \delta \;E_A} ) $$
represent a drop / flattened circle shapes of maximum heights as shown.
We can find intersections of these flattened circles... like graphically shown in the following or numerically from analytical form:
$$r_A \cos \theta_A-r_B \cos \theta_B= AB,\; r_A \sin \theta_A=r_B \sin \theta_B;\;$$
$$ \theta_A = \gamma,\; \theta_B= \pi- \delta\;$$

The construction shows heights $ (1,1.5) $ of drop shapes cutting at two points and corresponding  four polar angles on common base $( AB=1.25)$
Asking for all solutions is okay.. we should see what is relevant to the Physics and discard what is not.
