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.
Best Answer
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.