Right. Made a right balls up of the above, I've had another go...

So, A and B are 100km apart. The ISS is directly above A, B measures an angle of 76 degrees from the ground to the ISS. They assume that the ground between them is flat which forms a triangle. A-B is the base, 100km. The side of the triangle is going straight up from A to the ISS so it's a right angled triangle. The internal angle of the triangle A-B-ISS is 76 degrees which makes the other internal angle 14 degrees. Putting the known values into a triangle calculator you get a height of 401km

Actually though, A and B are on a sphere and although they are 100km apart they are 1 degree apart on the circumference of the sphere. So the total circumference is 36000km

(360 degrees in a circle, 100km is 1 degree)

Find the radius:

C = 2 pi r

so r = C / 2 pi

36000 / 2 pi = 5729.58km

Chord length is given by 2r sin (a/2) - where r is the radus of the circle, a is the angle at the centre of the circle (angle ACB above). That is 1 degree, so:

2 x 5729.58 x sin(0.5) = 99.998

So this is the straight line length between A and B through the curve of the circle.

The triangle ABC is isosoles so if the angle ACB is 1 degree then CAB and CBA must both be 89.5 degrees. That means there is 0.5 degrees between the chord and the tangent to the circle at both ends which we must adjust for. So:

A observes the ISS directly above him but adding the adjustment makes the angle

ISS-A-B 90.5 degrees

B measures an angle of 76 degrees, adding the adjustment that makes the angle

A-B-ISS 76.5 degrees.

The straight line AB is 99.998 so now we cancalculate the other sides of the triangle.

So...I'll be honest, this isn't what I was expectint but now the calculated height is 432km. I'll admit this is a much bigger error than I was expecting.

Have I done something wrong or is this just how it is?