Analog of Heron’s formula on a sphere

The area of a triangle can be computed directly from the lengths of its sides via Heron's formula.

Heres is the semiperimeter,s = (a +b +c)/2.

Is there an analogous formula for spherical triangles? It's not obvious there should be, but there is a formula by Simon Antoine Jean L'Huilier (1750-1840).

Here we denote area by S for surface area, rather than A because in the context of spherical trigonometry A usually denotes the angle opposite side a. The same convention applies in plane trigonometry, but the potential for confusion is greater in L'Huilier's formula since the area appears inside a tangent function.

Now tan for small , and so L'Huilier's formula reduces to Heron's formula for small triangles.

Imagine the Earth as a sphere of radius 1 and take a spherical triangle with one vertex at the north pole and two vertices on the equator 90 longitude apart. Then a = b = c = /2 and s = 3/4. Such a triangle takes of 1/8 of the Earth's surface area of 4, so the areaS is/2. You can verify that L'Huilier's formula gives the correct area.

It's not a proof, but it's a good sanity check that L'Huilier's formula is correct for small triangles and for at least one big triangle.

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