# Deriving the Surface Area of a Spherical Triangle

A triangle on a sphere is composed of points $A$, $B$ and $C$. The $\alpha$, $\beta$ and $\gamma$ denote the angles at the corresponding points of the triangle:

The Girard's theorem states that the surface area of any spherical triangle:

$$A = R^2 \cdot E$$

where $R$ is the radius of the sphere and $E$ is the excess angle of $(\alpha + \beta + \gamma - \pi)$

I'm wondering how to derive this formula.

• – Inquest Feb 16 '12 at 18:03
• TeX tip: Notice the difference between $\alpha$ + $\beta$ + $\gamma$ - $\pi$ and $\alpha + \beta + \gamma - \pi$. In the former, the minus sign looks like a hyphen instead of a minus sign. There's a reason why that's not standard usage. I fixed this in the posting. – Michael Hardy Feb 16 '12 at 18:06
• @Nunoxic, no. That question is about how to determine the angles between ABC, ACB and BAC. – ezpresso Feb 16 '12 at 18:06
• There is a nice proof at math.rice.edu/~pcmi/sphere/gos4.html – Michael E2 Jan 24 '15 at 22:37

Consider the following three parts of the sphere: let $$P_A$$ be the lune created from the triangle $$ABC$$ plus the triangle adjacent across the $$BC$$ line segment, plus the opposite lune (on the opposite side of the sphere), and similarly for parts $$P_B$$ and $$P_C$$.

The area of $$P_A$$ is $$4\alpha R^2$$: the total area of the sphere is $$4\pi R^2$$, and the area of $$P_A$$ is certainly proportional to $$\alpha$$.

Notice now that $$P_A\cup P_B\cup P_C$$ is the entire sphere, and that $$P$$'s intersect at the triangle + the opposite triangle. We thus have:

area($$P_A$$) + area($$P_B$$) + area($$P_C$$) = area of the sphere + 2 area of the triangle +2 area of the opposite triangle.

As the two triangles have the same area, you get your formula.

• You really should explain what they cut the circle to 4 parts actually means. The first part is the made up of the lunes $ABA'C$ and $A'B'AC'$ put together, for example. Otherwise this is a good argument by symmetry. – anon Feb 16 '12 at 19:53
• @anon: I meant "they cut the sphere"; thanks for spotting it! – user8268 Feb 16 '12 at 19:57
• I've edited it to reflect what I believe you were saying. Additionally, if we regard $A'$, $B\,'$ and $C\,'$ respectively as the points opposite to $A$, $B$ and $C$ on the sphere, we have the decomposition $$\begin{array}{} P_A=ABA'C\;\cup\;A'B\,'AC\,' \\ P_B= BCB\,'A\;\cup\; B\,'C\,'BA \\ P_C= CAC\,'B\;\cup\; C\,'A'CB\,' \end{array}$$ – anon Feb 16 '12 at 20:18
• $P_B = BCB'A \cup B'C'BA'$? – ezpresso Feb 16 '12 at 21:02
• @Seankala each $P_i$ is a pair of opposite lunes and they indeed cover the entire sphere, and the triangle (and also the opposite triangle) is covered 3 times (the rest is covered just once). I should have drawn a picture, I guess :) – user8268 Apr 1 '20 at 16:59

A bit late to the party, but we can also use Gauss-Bonnet theorem, which can be used for other $2$-dimensional manifolds as well, e.g. the hyperbolic plane. It states that if $M$ is a $2$-dimensional Riemannian manifold with piecewise smooth boundary, then $$\int_M K dA +\int_{\partial M} k_g+\sum_i \alpha_i=2\pi \chi(M)$$ where $K$ is the Gauss curvature, $k_g$ the geodesic curvature (which is zero for geodesics), $\alpha_i$ the angles by which the tangent to the boundary turns at points where $\partial M$ is not smooth, and $\chi(M)$ the Euler characteristic of $M$.

In our case $M$ is the triangle, $K$ is identically $\frac{1}{R^2}$, the angles $\alpha_i$ are $\pi-\alpha,\pi-\beta,\pi-\gamma$ and $\chi(M)=1$. Therefore the formula becomes $$\frac{A}{R^2}+(\pi-\alpha+\pi-\beta+\pi-\gamma) = 2\pi\implies \boxed{A= (\alpha+\beta+\gamma-\pi)R^2}.$$

Reference : http://planetmath.org/areaofasphericaltriangle 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

SABC=(∠A+∠B+∠C−π)R2. (1) Incidentally, this formula shows that the sum of the angles of a spherical triangle must be greater than or equal to π, 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 π.

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′B′C′ 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′ are shown below.

At each vertex, these diangles form an angle of ∠A. Similarly, we can form diangles with vertices on the diameters BB′ and CC′ respectively.

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

Ssphere=2SAA′+2SBB′+2SCC′−4SABC. (2) Clearly, a diangle occupies an area that is proportional to the angle it forms. Since the area of the sphere is 4πR2, the area of a diangle of angle α must be 2αR2.

Hence, we can rewrite equation (2) as

4πR2=2R2(2∠A+2∠B+2∠C)−4SABC,
∴ SABC=(∠A+∠B+∠C−π)R2,
which is the same as equation (1). ∎

Let us make this a bit simpler. Let $T$ represent the triangle area, $A$ the lune defined by $\alpha$, $B$ the lune defined by $\beta$, and $C$ the lune defined by $\gamma$. Then $(A-2T)+(B-2T)+(C-2T)+2T=4R^2\pi$. Also $A=4R^2\alpha$, $B=4R^2\beta$, and $C=4R^2\gamma$. Combining these areas gives $4R^2(\alpha+\beta+\gamma)-4T=4R^2\pi$ or $T=R^2(\alpha+\beta+\gamma-\pi)$.