## You are here

Homearea of a spherical triangle

## Primary tabs

# 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

$S_{{ABC}}=(\angle A+\angle B+\angle C-\pi)R^{2}.$ | (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 $AA^{{\prime}}$ are shown below.

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

$S_{{\mathrm{sphere}}}=2S_{{AA^{{\prime}}}}+2S_{{BB^{{\prime}}}}+2S_{{CC^{{% \prime}}}}-4S_{{ABC}}.$ | (2) |

Clearly, a diangle occupies an area that is proportional to the angle it forms. Since the area of the sphere is $4\pi R^{2}$, the area of a diangle of angle $\alpha$ must be $2\alpha R^{2}$.

Hence, we can rewrite equation (2) as

$\displaystyle 4\pi R^{2}=2R^{2}(2\angle A+2\angle B+2\angle C)-4S_{{ABC}},$ | ||

$\displaystyle\therefore~{}S_{{ABC}}=(\angle A+\angle B+\angle C-\pi)R^{2},$ |

which is the same as equation (1). ∎

## Mathematics Subject Classification

51M25*no label found*51M04

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Corrections

noncollinear by Wkbj79 ✓

Contains own proof by rm50 ✓

theorem pro definition by pahio ✓