area of a spherical triangle

A spherical triangle is formed by connecting three points on the surfaceof 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 sphericaltriangle is intuitively obvious, since on a small scale the surface ofa sphere looks flat. More precisely, the angle at each vertex is measured as theangle 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 sphericaltriangle must be greater than or equal to $\pi$, with equality holdingin case the triangle has zero area.

Since the sphere is compact, there might be some ambiguity as to whetherthe area of the triangle or its complement is being considered. Forthe purposes of the above formula, we only consider triangles witheach angle smaller than $\pi$.

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

Note that by continuing the sides of the original triangle into fullgreat 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 originalone through the center of the sphere. By symmetry, both triangles musthave the same area.

Proof.

For the proof of the above formula, the notion of a spherical diangleis helpful. As its name suggests, a diangle is formed by two great arcsthat intersect in two points, which must lie on a diameter. Two diangleswith 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 overlappingonly on the triangles $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$. Hence, the total area ofthe sphere can be written as

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

Clearly, a diangle occupies an area that is proportional to the angleit 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 $2\alpha R^2$.

Hence, we can rewrite equation (2) as

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

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

which is the same as equation (1).∎