# How to find the area of a spherical equilateral triangle?

Suppose an equilateral triangle is drawn on the surface of the earth (considered to be an exact sphere). Length of each side of the triangle is $L = 1$ km. The radius of the earth is $R = 6400$ km. How do you calculate the area of the triangle?

Is there any way to find the area of the triangle from the metric of the space, which is given by, in this case,

$ds^2 = R^2 \left( d\theta^2 + \sin^2 \theta \hspace{1mm} d\phi^2 \right)$.

There is a relation between the angles ($\alpha$, $\beta$ and $\gamma$) of the triangle on a surface with positive curvature, which is given by,

$\alpha + \beta + \gamma = \pi + \frac{A}{R^2}$

where, $A$ is the area of the triangle, $R$ is the radius of curvature of the surface and the angles are in radian. If we can find $A$, we can calculate the sum of the angles of a triangle on a curved surface.

## migrated from physics.stackexchange.comMar 29 '16 at 14:28

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• I think this a maths question rather than a physics question. – John Rennie Mar 29 '16 at 6:38
• You can always scale things to keep the constant unity. – GRrocks Mar 29 '16 at 6:44

Just like in Euclidean geometry, there exists a law of cosines in spherical geometry, for triangles on a unit sphere: $$\cos a = \cos b\cos c + \sin b\sin c\cos\alpha\\ \cos b = \cos c\cos a + \sin c\sin a\cos\beta\\ \cos c = \cos a\cos b + \sin a\sin b\cos\gamma$$ Note that the 'sides' $a,b,c$ of a spherical triangle are also angles. In your case, $a=b=c=L/R$ (in radians), from which you can calculate $\alpha=\beta=\gamma$, and subsequently the area.
Imagine the triangle is situated at the North Pole so that the corners are at $(\theta,\phi) = (0,0), (\theta_T,0)$ and $(\theta_T,\phi_T)$. From the metric $\mathrm d s^2 = g_{ij}\mathrm d x^i \mathrm d x^j = R^2(\mathrm d\theta^2+\sin^2\theta\mathrm d\phi^2)$ it is clear that we must have $\theta_T = \frac{L}{R}$ and $\phi_T=\frac{L}{R\sin\theta_T}$. Then the area can be calculated in the usual manner on a curved manifold
$$A = \int_{\mathrm{triangle}}\mathrm{d}^2x \sqrt{|\mathrm{det}\: g|} = R^2\int_0^{\phi_T}\mathrm{d}\phi\int_0^{\theta_T}\mathrm{d}\theta\sin\theta = \frac{RL}{\sin\frac{L}{R}}\left(1-\cos\frac{L}{R}\right) \simeq \frac{L^2}{2}\left[1+\frac{1}{12}\left(\frac{L}{R}\right)^2 + \mathcal O\left(\frac{L^4}{R^4}\right)\right].$$