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Pythagoras on a Sphere

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

On a sphere of radius $R$ we use a scale factor and the equivalent formula is $$\cos {a\over R} = \cos {b\over R} \cos {c\over R}.$$
All the familiar trigonometric identities in Euclidean Plane Geometry have their counterparts in Spherical Geometry (otherwise known as Elliptic Geometry) and also in Hyperbolic Geometry. These are the geometries on flat surfaces (Euclidean Geometry), on surfaces of positive curvature (Spherical Geometry) and on surfaces of negative curvature (Hyperbolic Geometry). See the article Curvature of Surfaces.

Most of the results in these two other geometries are much more similar in form to the trigonometric identities you meet in school than the result proved here which is equivalent to Pythagoras' Theorem.

The corresponding Pythagorean Theorem for right-angled triangles in Hyperbolic Geometry is: $$\cosh a = \cosh b \cosh c.$$