The area of a triangle can be computed directly from the lengths of its sides via Heron’s formula.
Here s 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.
Now tan θ ≈ θ for small θ, 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 = π/2 and s = 3π/4. Such a triangle takes of 1/8 of the Earth’s surface area of 4π, so the area S is π/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.