Find all triangles with a fixed base and opposite angle

I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.

I also know the angle opposite the base and this is also fixed.

Now what I want to do is figure out how to compute all possible positions for the third vertex.

My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.

I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis

then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste

now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.

struggling with the maths, do i use the sin rule, i.e. sin a / $=\mathrm{sin}b/B$ and so on, or do I need to break it down into right angle triangles and then just use something along the lines of ${a}^{2}+{b}^{2}={c}^{2}$

ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,

base is fixed running from (-1,0) to (1,0) angle opposite the base is 60 deg

I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.

I also know the angle opposite the base and this is also fixed.

Now what I want to do is figure out how to compute all possible positions for the third vertex.

My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.

I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis

then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste

now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.

struggling with the maths, do i use the sin rule, i.e. sin a / $=\mathrm{sin}b/B$ and so on, or do I need to break it down into right angle triangles and then just use something along the lines of ${a}^{2}+{b}^{2}={c}^{2}$

ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,

base is fixed running from (-1,0) to (1,0) angle opposite the base is 60 deg

I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.