200 km of an orbit are mostly a straight line, therefore it's extraneous.
Station A: (0, 0)
Station B: (200, 0)
Build two linear equations (rem. f(x) = mx + t)
m = dy/dx = tan(n)
=> First lin. equation of Station A: f(x) = tan(23°)*x
Second lin. equation of Station B: g(x) = tan(55°)*(x-200)
Solve f = g for x
tan(23°)*x = tan(55°)*(x-200)
=> tan(23°)/tan(55°) = (x-200)/x = 1-200/x
=> x = 200/[1-tan(23°)/tan(55°)]
Insert x into f(x):
f(200/[1-tan(23°)/tan(55°)]) = tan(23°)*200/[1-tan(23°)/tan(55°)] = 120.8 km
This would result into:
Another solution would occur, if you set the angle of g(x) to 180-55°. Station B would face the opposite direction.
=> f(200/[1-tan(23°)/tan(125°)]) = tan(23°)*200/[1-tan(23°)/tan(125°)] = 65.44 km
Which would lead to:
BUT a satellite in about 65 km height is unrealistic, so it has to be the first one.
+104 
