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Bill wrote-

>> For example, if you mark three arbitrary points on the equator,
>> you obtain a triangle whose three angles are 180 deg each.
>
>Alex
>
>I am having trouble finding a frame of reference where I can imagine three
>180d angles.  If I recall, Herbert or George also surmised it you example
>might be considered a total of 180d.
>
>From my perspective, if I raise the center point just a hair off the equator
>I would see a triangle.  The center angle close to 180d, and the other two
>close to 0d.  Same if I raise and end point just a hair.
>
>How do I adjust my frame of reference to imagine three 180d angles in your
>above example?
>
>Bill

==============

Mark three points around the equator. Point A is at 0deg West, point B is
at 120deg West, point C is at 240deg West. Join A to B, B to C, and C to A
again, with great circles, going Westerly each time. Then you have three
vertexes, each subtending a 180deg angle. The resulting triangle divides
the circle into two equal halves

You might object that with such a 180deg angle, then each vertex has become
a straight line, not a corner. That's true, because it's a limiting case.
Think about it, if you prefer, when the angle at A, B, and C is not quite
(but as near as dammit) equal to 180, so there's a VERY obtuse angle at
each corner. Then increase these angles, very slightly.

This business of the "spherical excess" is quite new to me, but it seems to
work. If we add the three "angles", each 180, we get 540 deg. The spherical
excess over a tiny triangle, in which the angles always sum to 180deg, is
therefore 360 deg, or 2pi radians. Multiply this by r squared, and we get 2
pi r-squared, which is indeed the area of the half-sphere that the
"triangle" embraces. Today, I've learned something new...

George.

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