Fig 3a shows a spherical triangle **I** with angles abc. This triangle is part of the lune with poles R and R' (3b). This lune consists of spherical triangles **I** + **IV** (3c). At pole R, we can see the lune has angle c. Therefore the area of the lune with poles R and R' = 2R²c.

**I** + **IV** = 2R²c.

By the same method, the lune with poles Q and Q' consists of spherical triangles **I** + **II**. So the area of this lune is:

**I** + **II** = 2R²b.

And the area of the lune with poles P and P' is:

**I** + **III'** = 2R²a.

Note that in 3c, each spherical triangle at the front **I**, **II**, **III**, **IV**, has a diametrically opposite triangle at the back **I'**, **II'**, **III'**, **IV'**. Thus **III** and **III'** are congruent, which means the last equation can be rewritten as:

**I** + **III** = 2R²a.

Adding the 3 equations together gives:

3 • **I** + **II** + **III** + **IV** = 2R²(a + b + c)

It is obvious that **I** + **II** + **III** + **IV** is a hemisphere:

**I** + **II** + **III** + **IV** = 2R²

And by symmetry about the center, **I** + **II** + **III** + **IV'** is also a hemisphere because **IV'** is congruent and diametrically opposed to **IV**.

**I** + **II** + **III** + **IV** = **I** + **II** + **III** + **IV'** = 2R²

Note: this applies this to any congruent pair of spherical triangles. For example, in the case of **II** and **II'**, we can say:

**I** + **II** + **III** + **IV** = **I** + **II'** + **III** + **IV** = 2R²

Thus,

2 • **I** + 2R² = 2R²(a + b + c)

So

**I** = R²(a + b + c - ) = (R² / 180°)(a° + b° + c° - 180°)

Therefore, the area of **I** is:

**I** = R² [ (a + b + c) - ]

The area A of a spherical triangle is:

**A** = R² [ (a + b + c) - ] (if a, b and c are in radians).

**A** = R² [ (a° + b° + c°) - 180° ] (if a, b and c are in degrees).

It has already been shown that the sum of the angles of a spherical triangle are together greater than two right angles (). The difference between this sum and (or 180°) is called the spherical excess.

The area of a spherical triangle is equal to its spherical excess.

[Ref: ‘The VNR Concise Encyclopedia of Mathematics’ p.263. W. Gellert, H. Küstner, M. Hellwich, H.Kästner (K.A Hirsch, H. Reichart), Bibliographisches, Institut Leipzig 1975]

[Ref: ‘Geometry’ p.341-342. D. A. Brannan, M. F. Esplen, J. J. Gray, Cambridge University Press]