Hello Archie Meade -

Sorry to pursue this further, but what you (and others) don't seem to be taking into account is that the distance of the earth from the moon - given as 240,000 miles - is:

(a) very large compared to the diameter of the moon, and, for that matter, the earth;

(b) only given to 2 significant figures, anyway. (The mean distance is given on WikiAnswers as 238,857 miles.)

Thus, taking the moon's radius into account is a waste of time and effort. You might just as well consider the radius of the earth as well.

The fact that the distance between the earth and the moon is large compared to their radii is the reason that I was able to offer the solution that I did. Because of this fact, I was able to treat the visible surface of the moon as an arc of a circle, whose centre is the observer on the surface of the earth.

Captain Black's solution (using the tangent of half the subtended angle) assumes that the distance of 240,000 miles is the distance from the observer to the centre of the moon. But equally, you could take it (with no reasonable loss of accuracy) as the distance from the observer to the visible edge of the moon, in which case the tangent of the angle is replaced by its sine.

skeeter's solution, using the Cosine Rule, assumes the same thing.

The fact is that for small values of (in radians). So the difference between the arc length (which was the method I used) and the length of the chord is totally irrelevant compared to the other factors I have listed above.

So rather than offer more complicated solutions that add nothing of value, please let's keep it simple!

Grandad