Compute the area of a triangle on a surface of the earth which has a spherical excess of 1.75″. Assume radius of curvature as 6400000m. Express the area in square kilometers.
347.51sqkm.
275.316sqkm.
414.325sqkm.
137.425sqkm.
\[
e” = \frac{AREA}{R^2 \sin 01″}
\]
\[
1.75 = \frac{AREA}{(6400000m.)^2 \sin 01″}
\]
\[
AREA = 1.75{(6400000m.)^2 \sin 01″}
\]
\[
AREA = 347514446.60sqm. \frac{(1km.)^2}{(1000m.)^2}
\]
\[
AREA = 347.5144466sqkm.
\]
\[
AREA \approx 347.51sqkm. \leftarrow \text{answer}
\]