Exercise 2.8.  Show that given a sphere with radius R, the surface area of a triangle T with vertices A, B, C and interior angles α, β, and γ inscribed on the surface of the sphere is R²(α+β+γ-π).

     We can start about by examining the surface area swept out by a single pair of great circles forming one of the interior angles of T.  For instance, the surface area swept out by the pair of great circles intersecting to form angle β:

     When β equals π, the area swept out covers the entire surface area of the sphere, or 4πR².  Therefore we can infer (dubious, see below discussion) the following relationship between the great circles intersecting to form angle β and the surface area A_β swept out by these great circles:

A_β = 4βR²

(1)

    

     Armed with this knowledge, let's look at all of the surface areas swept out by the great circles intersecting to form all of T's interior angles:

     Using equation (1), the total surface area swept out by the great circles intersecting to form the three interior angles for T is:

A_α,β,γ = 4R²(α+β+γ)

     Then note that the area of T, A_T, is swept out three times by each pair of great circles, and is mirrored on opposite poles of the sphere, so that triangle ABC is congruent to triangle A'B'C' on the sphere.  Therefore A_α,β,γ covers the entire surface area of the sphere plus extra area equal to a multiple of A_T:

4R²(α+β+γ)=4πR²+4A_T

       (2)

      

     Four times A_T because A_T is swept out two "extra" times on each pole of the sphere.  Using (2) it is easy to solve for the correct value of A_T.  (2) also indicates that the sum of the angles of a triangle inscribed on the surface of a sphere must always be greater than π.

Dubiousness...

      I'm not particularly happy with the way I just kinda closed one eye and went ahead with "inferring" expression (1) for the surface area swept out by the great circle intersections.  It seems right but I dunno whether there's something more I'm supposed to be doing so that I could nail that down with more rigor.  Or maybe that's par for the course given that Road is not after all a full-blown math course.  

     Anyway I got to the right answer without too much B.S. so I guess I should count myself lucky.  Working to finish up a beast right now that I may have leave unfinished and ask for help from whoever the hell might eventually one day actually read this stupid blog.  (hi, unlucky historian of the Internet reading this a thousand years in the future!)