# area of a spherical triangle

A spherical triangle is formed by connecting three points on the surface
of a sphere with great arcs; these three points do not lie on a great circle of the sphere. The measurement of an angle of a spherical
triangle is intuitively obvious, since on a small scale the surface of
a sphere looks flat. More precisely, the angle at each vertex is measured as the
angle between the tangents^{} to the incident^{} sides in the vertex tangent plane^{}.

Theorem. The area of a spherical triangle $ABC$ on a sphere of radius $R$ is

$${S}_{ABC}=(\mathrm{\angle}A+\mathrm{\angle}B+\mathrm{\angle}C-\pi ){R}^{2}.$$ | (1) |

Incidentally, this formula^{} shows that the sum of the angles of a spherical
triangle must be greater than or equal to $\pi $, with equality holding
in case the triangle^{} has zero area.

Since the sphere is compact, there might be some ambiguity as to whether
the area of the triangle or its complement^{} is being considered. For
the purposes of the above formula, we only consider triangles with
each angle smaller than $\pi $.

An illustration of a spherical triangle formed by points $A$, $B$, and $C$ is shown below.

Note that by continuing the sides of the original triangle into full
great circles, another spherical triangle is formed. The triangle ${A}^{\prime}{B}^{\prime}{C}^{\prime}$
is antipodal to $ABC$ since it can be obtained by reflecting the original
one through the center of the sphere. By symmetry^{}, both triangles must
have the same area.

###### Proof.

For the proof of the above formula, the notion of a *spherical diangle*
is helpful. As its name suggests, a diangle is formed by two great arcs
that intersect in two points, which must lie on a diameter^{}. Two diangles
with vertices on the diameter $A{A}^{\prime}$ are shown below.

At each vertex, these diangles form an angle of $\mathrm{\angle}A$. Similarly, we can form diangles with vertices on the diameters $B{B}^{\prime}$ and $C{C}^{\prime}$ respectively.

Note that these diangles cover the entire sphere while overlapping only on the triangles $ABC$ and ${A}^{\prime}{B}^{\prime}{C}^{\prime}$. Hence, the total area of the sphere can be written as

$${S}_{\mathrm{sphere}}=2{S}_{A{A}^{\prime}}+2{S}_{B{B}^{\prime}}+2{S}_{C{C}^{\prime}}-4{S}_{ABC}.$$ | (2) |

Clearly, a diangle occupies an area that is proportional to the angle it forms. Since the area of the sphere (http://planetmath.org/AreaOfTheNSphere) is $4\pi {R}^{2}$, the area of a diangle of angle $\alpha $ must be $2\alpha {R}^{2}$.

Title | area of a spherical triangle |

Canonical name | AreaOfASphericalTriangle |

Date of creation | 2013-03-22 14:21:38 |

Last modified on | 2013-03-22 14:21:38 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 9 |

Author | Mathprof (13753) |

Entry type | Theorem |

Classification | msc 51M25 |

Classification | msc 51M04 |

Related topic | AreaOfTheNSphere |

Related topic | Defect |

Related topic | SolidAngle |

Related topic | LimitingTriangle |

Related topic | SphericalTrigonometry |