>You need three points in a triangulation problem yes: target,

>observation point 1, observation point 2. We've only got observation

>point 1's location as a known value. Actually, for triangulation in 2d space, you need 3 observation points. One gives you a circle, two gives you two points (intersection of two circles), and the third tells you which of those two points is the right one. You can probably get away without the 3rd, if there's any directionality to the signal at all...it's only if it's omni-directional that you'd need them all.

Likewise, in 3d space, you'd need four. One gives a sphere, two gives a circle (intersection of the surfaces of two spheres), three gives two points (intersection of the surface of a sphere and a circle), and fourth tells you which of the two points.

and goodness, geometry was so many years ago...