Girard's theorem
/ zhee-RAR /
How big is a triangle drawn on a sphere? On a flat plane you would need the base and height; on a sphere there is a far more surprising answer, and Girard's theorem (named for Albert Girard, with Thomas Harriot finding it earlier) gives it. The area of a spherical triangle depends ONLY on its angles. Precisely, on a sphere of radius R, area = R^2 * (A + B + C - pi), where A, B, C are the three angles in radians. The quantity (A + B + C - pi) is the angle excess; so area equals R^2 times the excess — a clean, exact law.
Here is the cleverest way to see why. A 'lune' is the orange-slice wedge between two great circles, and a lune of angle A has area 2 * R^2 * A (it is the fraction A/(2 pi) of the whole sphere's area 4 pi R^2, doubled because a lune spans pole to pole twice over). Now a triangle's three sides, extended to full great circles, carve the sphere into the triangle, its mirror antipodal copy, and three pairs of lunes. Adding the three lunes' areas double-counts the triangle in a controlled way, and the bookkeeping collapses exactly to area = R^2 (A + B + C - pi). It is one of the most satisfying short proofs in all of geometry.
The theorem's reach is large. It is the spherical sibling of the hyperbolic law area = R^2 (pi - A - B - C), and both are baby cases of the grand Gauss-Bonnet theorem linking total curvature to topology. Practically, it lets surveyors and astronomers compute areas on a globe from angle measurements alone, and it shows there is a largest spherical triangle (excess capped at the area of a hemisphere). The honest caveat is the same as for the sphere itself: because great circles meet twice, this lives most cleanly on the sphere, with elliptic geometry obtained by identifying antipodes.
Take the triangle with one vertex at the North Pole and two on the equator, 90 degrees of longitude apart, so A = B = C = 90 degrees = pi/2. Then A + B + C - pi = 3 pi/2 - pi = pi/2, and the area is R^2 * pi/2. Check: this triangle is exactly one-eighth of the sphere, whose total area is 4 pi R^2, and one-eighth of that is indeed pi R^2 / 2.
Area = R^2 (A + B + C - pi): a spherical triangle's area is just R^2 times its angle excess, proven by adding three lunes.
The astonishing content is that side lengths never appear — angles alone fix the area, the spherical mirror of the hyperbolic defect-area law. Remember to convert the angle excess to radians before multiplying by R^2.