The interior angles in triangle ABC are A=57°30'29", B=65°17'27", and C=57°12'16". The distance from A to B has been computed to be 44386.45m. Assuming the average curvature of 6400km., determine the spherical excess in the triangle. 4.52" 3.86" 5.12" 2.82" Click to Show/Hide Solution \[ e" = \frac{AREA}{R^2 \sin 01"} \] \[ \frac{a}{\sin A} = \frac{c}{\sin C} \] \[ a = \frac{c (\sin A)}{\sin C} \] \[ a = \frac{44386.45m. (\sin 57^{\circ}30'29")}{\sin 57^{\circ}12'16"} \] \[ a = 44537.38m. \] \[ AREA = \frac{ac \sin B}{2} \] \[ AREA = \frac{(44537.38m.)(44386.45m.) \sin 65^{\circ}17'27"}{2} \] \[ AREA = 897928915.60sqm. \] \[ e" = \frac{897928915.60sqm.}{{6400km.(\frac{1000m.}{1km})}^2 \sin 01"} \] \[ e" = 4.52 \leftarrow \text{answer} \]