A spherical triangle is a figure on the surface of a sphere, bounded by three arcs of great circles. The shape is fully described by six values: the length of the three sides (the arcs) and the angles between sides taken at the corners. If the side lengths are given in angular measure, the shape description becomes independent of the sphere's radius. This Demonstration solves the unknown three values when the other three are known.
Each side arc defines a plane, which also contains the sphere center. The corner angles are equal to the angles between these planes. The angle between two planes is called a dihedral angle.
There are several conventions for labeling corners, sides, and their angles. Here the corner angles are labeled with capital letters

,

, and

. The angles of opposite sides are labeled

,

, and

.
Given three corner or side angles, a solution may or may not exist. If a solution exists it may not be unique. The number of solutions, if any, is either one, two, or infinitely many.
Closely associated with each spherical triangle is its trihedron. In this context it is the triple of lines from the sphere's center to the triangle corners. In other words, the trihedron is the set of position vectors of the corners, with the origin at the sphere's center. An alternative definition of the trihedron also includes the three faces bounded by the lines and triangle sides. The Demonstration visualizes both definitions.
The area of the interior of the triangle can be calculated from the corner angles:

, where the angles are expressed in radians.
The most common conventions restrict a spherical triangle to have all corners and sides less than 180°. In contrast this Demonstration allows angles up to and including 180°. Thereby a spherical triangle can take the shape of a spherical lune.
To apply this Demonstration to great circle navigation, mentally model the corner

to be located at the north pole and the location of the corners

and

to points of departure and arrival. At the corner angle

input the difference in longitude between the locations

and

. Input the colatitudes of the locations

and

as side angles

and

, respectively. The solution gives the great circle distance in nautical miles as the side angle

times 60. The course steered at departure is given by the corner angle

and the course steered at arrival is the supplement to the corner angle

.