I have two points and the distance between them.
I'm converting them from lat/lon to cartesian coordinates.
I have the distance between A and B (represented by x) and I want to find C based on the following conditions:
the angle a must be between 50- 150 degrees.
the distance between point A and B should be between 1.7x and 30x.
Just for understanding the problem, I'm writing a script (in python) that needs to extract the range of all those points.
And I'm not sure what difference does it make, but it is a 3d triangle.
That’s a rather broad range of values. For any choice of values for the angle $\alpha$ and ratio $AB:AC$, there will be an infinite number of solutions lying in a circle perpendicular to $\overline{AB}$. Even in 2-D, there will be two solutions, symmetric with respect to that line. The easiest thing to do, I think, would be to move everything to the $x$-$y$ plane, find one of these solutions, then map that back to the actual setup with an appropriate scaling, rotation and translation. The complete solution set can then be obtained by rotation about the $AB$ axis.
So, working in the plane, by the Inscribed Angle theorem, the point $C$ lies somewhere on the arc of a circle with endpoints $A$ and $B$ such that the arc subtends an angle of $2\alpha$. When $\alpha\gt\pi/2$, this point will be on the minor arc defined by $A$ and $B$, and the angle at the center will instead be $2(\pi-\alpha)$. Since $\overline{AB}$ is a chord of the circle, the center $D$ lies somewhere on its perpendicular bisector, and if $M$ is the midpoint of the chord, $\triangle{DMA}$ is a right triangle. At the same time, $C$ must also lie on a circle with center $A$ and radius $x$, so the problem reduces to intersecting a pair of circles.
W.l.o.g. we can take $AC=x=1$. Place $A$ at the origin and $B$ at $(2m,0)$, with $m>0$. The center of the first circle is then at $D=\left(m,m\cot\alpha\right)$. This will place the center above the $x$-axis for $\alpha\lt\pi/2$ and below it for $\alpha\gt\pi/2$. An equation of this circle is $$(x-m)^2+(y-m\cot\alpha)^2=m^2\csc^2\alpha.$$ The second circle is simply $x^2+y^2=1$. The general solution to this system isn’t pretty, but it is straightforward. The point $C$ will be the solution with the greater $y$-coordinate. Scale, rotate and translate appropriately to place this on the actual points $A$ and $B$.