The interior angles in triangle ABC are A=82°14’44”, B=47°39’54”, and C=50°07’34”. The distance from A to B has been computed to be 12383.42m. Assuming the average curvature of 6372160m., determine the spherical excess in the triangle.
- 0.37″
- 1.42″
- 0.68″
- 2.42″
\[
e” = \frac{AREA}{R^2 \sin 01″}
\]
\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\]
\[
a = \frac{c (\sin A)}{\sin C}
\]
\[
a = \frac{12383.42m. (\sin 82^{\circ}14’44”)}{\sin 50^{\circ}07’34”}
\]
\[
a = 15988.09m.
\]
\[
AREA = \frac{ac \sin B}{2}
\]
\[
AREA = \frac{(15988.09m.)(12383.42m.) \sin 47^{\circ}39’54”}{2}
\]
\[
AREA = 73178045.23sqm.
\]
\[
e” = \frac{73178045.23sqm.}{(6372160m.)^2 \sin 01″}
\]
\[
e” = 0.37 \leftarrow \text{answer}
\]
