The sum of the internal angles of a triangle on a sphere is greater than 180°, and related to the area of the triangle

The area of a spherical triangle with interior angles α, β, γ is S = r2 (α + β + γ - π), a remarkable formula because it does not contain the lengths of the sides. This formula is most easily found by using diangles (translation of the German Zweieck). It is axiomatic in the plane that two straight lines cannot contain an area. On the sphere, however, two straight lines -- that is, great circles -- define an area, which we shall call a diangle. The two sides of a diangle are half-circumferences of the sphere, and its two angles are also equal. The area of a diangle is the area of the sphere times the ratio of the angle of the diangle to 2π radians, or S = 2r2θ, where θ is the angle of the diangle. A diangle is the figure enclosed by two equal straight lines, which is what great circles are, on the sphere.

Now sketch a sphere and a spherical triangle upon it. Extend each side of the triangle completely around the sphere, noting that an equal triangle is formed on the diametrically opposite side of the sphere from the original triangle. What one really needs to draw diagrams for spherical geometry is a sphere one can draw upon, and a hemispherical base to make the drawing and measuring of great circles easy. If we had one of these, what is about to be described would be very easy to picture. The sketch will just do, with an effort to understand what is going on in three dimensions.

Consider any pair of sides. The extended sides form two diangles, covering part of the sphere. Now take another pair of sides. When extended, these form two more diangles that do not overlap the first two, except within the triangle and its doppelganger on the other side of the sphere. The final pair of sides make two diangles that complete the covering of the sphere, and again overlap the two triangles. Therefore, the sum of the areas of the six diangles is equal to the area of the sphere, plus the four triangle areas extra: 4r2 (α + β + γ) = 4πr2 + 4S. Hence, S = r2(α + β + γ - π), which was to be proved.

Imagine walking around the triangle on its sides. When you come to the corner with internal angle α, you must turn through an angle of π - α. By the time you return to your starting place, you have had to turn through a total angle of Δ = 3π - α - β - γ. The area of the triangle can easily be expressed in terms of this angle: S = r2(2π - Δ). The area of the triangle can, therefore, be expressed solely in terms of how far you turn in circumambulating it. For a triangle in a plane, Δ is always 2π radians, as you may find obvious, giving S = 0 in the limit, which is perfectly appropriate when you think about it, since we have 0·∞.

On the sphere, you do not have to turn through 2π radians in following a path that leads back to where you started. As Magellan proved, you only have to go far enough in a straight line to return to your starting point. In this case, Δ = 0, and S = 2πr2, the area of a hemisphere, which is indeed the area enclosed by your path. If you proceed a distance s along a path of curvature k (the reciprocal of the radius of curvature), you turn through an angle ks. Using integral calculus, this means that Δ = ∫ kds, or S = r2(2π - ∫ kds), which is the celebrated formula of Gauss and Bonnet. k is the geodesic curvature of the path, which is zero for a great circle, since a great circle is a straight line on a sphere. The case of abrupt corners can be handled by including an explicit term for them, or by adding (π - α)δ(sα - s), where δ(x) is the Dirac delta function, for a corner of internal angle α at position sα.

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Composed by J. B. Calvert
Created 1998
Put into HTML 19 November 2000