Finding Area of a Spherical Triangle

Date: 07/08/2004 at 20:32:13
From: Denise
Subject: spherical triangles

Assuming the earth is a sphere, and given three (latitude, longitude)
coordinates, what's the easiest way to calculate the area of the
spherical triangle formed by those three points?

For example: What is the area between coordinates 34, -105; 34.1,
-105.5; 34.2, -105.9 ?

I have found various formulas to calcuate spherical triangles, such as 
Girard's theorem, but they assume I know the angles.  I don't know how 
to go from coordinates to angles and take into account all the other 
stuff like great circles.

I tried finding the distances between the 3 points and then using 
1/4*square root (p*(p-2a)(p-2b)(p-2c)), where a,b, and c are the 
lengths of the sides and p is the perimeter, but of course this 
doesn't take into consideration that the Earth is not flat.

Date: 07/09/2004 at 14:07:56
From: Doctor Rick
Subject: Re: spherical triangles

Hi, Denise.

My CRC Standard Mathematical Tables contain the basic formula for the 
area of a spherical triangle:

  Area = pi*R^2*E/180

where

  R = radius of sphere
  E = spherical excess of triangle, E = A + B + C - 180
  A, B, C = angles of spherical triangle in degrees

This is the formula you say isn't helpful because you don't know the 
angles, right? Well, the tables also have the following formula for 
the spherical excess E:

  tan(E/4) = sqrt(tan(s/2)*tan((s-a)/2)*tan((s-b)/2)*tan((s-c)/2))

where

  a, b, c = sides of spherical triangle
  s = (a + b + c)/2

You can find the sides using either the cosine formula or the 
haversine formula, found on the following pages in the Dr. Math archives:

  Distance using Latitude and Longitude
    http://mathforum.org/library/drmath/view/54680.html 

  Deriving the Haversine Formula
    http://mathforum.org/library/drmath/view/51879.html 

Good luck!  Write back if you are still stuck and show me what you've
been able to do.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/