I'll deal just with the case of A = [itex]\frac{\pi}{2}[/itex], that is the velocity is at right angles to the line joining the charges. The easiest way to do this is to start in the frame of reference in which the charges are stationary. The force between them in this frame is simply the ordinary Coulomb's law force. Now transform this force to the lab frame, in which the charges are moving. The force transformation is very easy as
transverse force = change in transverse momentum / time taken to change.
The change in transverse momentum is the same in both frames, as it is a Lorentz invariant. The time is dilated by the usual gamma factor in the lab frame, compared with that in the frame in which the charges are at rest. So the force between the charges in the lab frame is, in SI units
[tex]\frac{\sqrt{1 - \frac{v^2}{c^2}} Q^2}{4 \pi \epsilon_0 d^2}.[/tex]
This reduction in the force can be interpreted as a magnetic (Ampère) attractive force coming into play in the opposite direction to the electric (Coulomb) repulsive force. But it's not as simple as that… As Zoki's equations show, the electric field is also changed. You'll note that Zoki's equations boil down to give the result I've derived above, in the special case of velocity at right angles to the line joining the charges. When the angle is not a right angle things get more complicated, but, again, you can either go for a relativistic force transformation approach (treating components parallel to, and transverse to, the velocity separately), or you can use Zoki's equations (having first derived them?)