A right spherical triangle ABC of a sphere of radius 50 cm has angle C = 90 deg, A = 56 deg and side b = 47 deg. Find the area of the triangle in square centimeter.

SIN-COOP Rule
$\sin \bar{B} = \cos \bar{A} \cos b$

$\sin (90^\circ - B) = \cos (90^\circ - 56^\circ) \cos 47^\circ$

$B = 55.57^\circ$
 

 

Spherical excess
$E = A + B + C - 180^\circ$

$E = 56^\circ + 55.57^\circ + 90^\circ - 180^\circ$

$E = 21.57^\circ$
 

$\text{Area} = \dfrac{\pi R^2 E}{180^\circ} = \dfrac{\pi(50^2)(21.57^\circ)}{180^\circ}$

$\text{Area} = 941.17 ~ \text{cm}^2$