Plane Geometry - Bruce E. Shapiro

SECTION 53. SPHERICAL GEOMETRY 317Consider the hemisphere formed by the spherical line ÃB. Since the wholesphere has area 4π, the hemisphere has area 2π, and it can be divided upinto four spherical triangles (we drop the S subscript here):2π = α(△ABC) + α(△A ′ BC)+α(△AB ′ C) + α(A ′ B ′ C)(53.1)Consider each of the lunes formed by the complements of each angle. Denotinga lune with central angle θ by L θ ,α(L π−A ) = α(△AB ′ C) + α(△A ′ B ′ C)α(L π−B ) = α(△A ′ BC) + α(△A ′ B ′ C)α(L π−C ) = α(△A ′ BC) + α(△AB ′ C)Adding the three equations,2[α(△A ′ BC) + α(△AB ′ C) + α(△A ′ B ′ C)]= α(L π−A ) + α(L π−B ) + α(L π−C )= 2(π − A) + 2(π − B) + 2(π − C)= 2(3π − A − B − C)(53.2)where we have used A, B and C as a short-hand to denote the measures ofthe respective angles.The left hand side of equation 53.2 is precisely twice the sum of the lastthree terms in equation 53.1, henceRearranging gives4π = α(△ABC) + 3π − A − B − Cα(△ABC) = π + A + B + CThe following is sometimes called the Pythagorean Theorem on a Sphere.Theorem 53.11 Let △ABC be a spherical right triangle on a sphere ofradius r with right angle at C. Let the arc lengths of the sides opposite A,B, and C be a, b, and c. Thencos c (r = cos a ) Åcos b ãr rRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.