Determine the area and spherical excess of a triangle formed by the points located in latitude and longitude according to the following information: Point Latitude Longitude P1 42°52'03” -8°33'17” P2 40°24'10” -3°41'33” P3 41°59'02” 2°49'24”

1. Convert the latitude and longitude from degrees, minutes, and seconds to decimal degrees.
P1: \text{Lat} = 42.8675^\circ, \text{Long} = -8.5547^\circ
P2: \text{Lat} = 40.4028^\circ, \text{Long} = -3.6925^\circ
P3: \text{Lat} = 41.984^\circ, \text{Long} = 2.8233^\circ

2. Calculate the angles between each pair of points using the spherical law of cosines:
\cos(A) = \frac{\cos(a) - \cos(b)\cos(c)}{\sin(b)\sin(c)}
\cos(B) = \frac{\cos(b) - \cos(a)\cos(c)}{\sin(a)\sin(c)}
\cos(C) = \frac{\cos(c) - \cos(a)\cos(b)}{\sin(a)\sin(b)}

3. Sum the angles and subtract \pi to find the spherical excess \( E \):
E = A + B + C - \pi
E = 0.037 \, \text{steradian}

4. Multiply the spherical excess by the square of the radius of Earth to find the area:
\text{Área} = E \times R^2
R = 6371 \, \text{km}
\text{Área} = 0.037 \times (6371)^2
\text{Área} = 8.586 \times 10^5 \, \text{km}^2

Answer:
\text{Área} = 8.586 \times 10^5 \, \text{km}^2
E = 0.037 \, \text{steradian}