Area of Spherical Triangle-2

Here is a presentation for detailed explanation on how to compute area of spherical triangle given all three sides

One can easily compute the interior angles of spherical triangle ABC using cosine formula

$A=\cos^{-1}\left(\frac{\cos\frac{5}{20}-\cos\frac{12}{20}\cos\frac{13}{20}}{\sin\frac{12}{20}\sin\frac{13}{20}}\right)=0.420997306$

$B=\cos^{-1}\left(\frac{\cos\frac{12}{20}-\cos\frac{5}{20}\cos\frac{13}{20}}{\sin\frac{5}{20}\sin\frac{13}{20}}\right)=1.201819972$

$C=\cos^{-1}\left(\frac{\cos\frac{13}{20}-\cos\frac{5}{20}\cos\frac{12}{20}}{\sin\frac{5}{20}\sin\frac{12}{20}}\right)=1.59652762$

hence area of spherical triangle ABC

$=(A+B+C-\pi)R^2=(0.420997306+1.201819972+1.59652762-\pi)20^2=31.10089784 \ cm^2$

Using the cosine rule from spherical trigonometry, which is as follows. For a spherical triangle with sides $a, b, c$ (which are the angles the sides subtend at the center of the sphere), we have the following rule:

$\cos a = \cos b \cos c + \sin b \sin c \cos A$

$\cos b = \cos a \cos c + \sin a \sin c \cos B$

$\cos c = \cos a \cos b + \sin a \sin b \cos C$

Where $A, B, C$ are the spherical angles at the three vertices of the spherical triangle, with angle $A$ between sides $b$ and $c$ , and so on. The above equations determine the desired angles $A, B , C$ . What remains is to use the formula for the area of a spherical triangle, which is,

$\text{Area} = R^2 (A + B + C - \pi )$

Applying the cosine rule above with $a = \dfrac{5}{20} , b = \dfrac{12}{20}, c = \dfrac{13}{20}$ , we obtain,

$A = 0.420997306 , B =1.201819972 , C = 1.59652762$ . Therefore,

$\text{Area} = (20)^2 ( 0.420997306 + 1.201819972 + 1.59652762 - 3.141592654 ) = 31.10089801$