P]AOB is equal to the interior angle P of the spherical triangle
Consider again a spherical triangle with vertex C at the North Pole (Fig.
Table 1 The angles in a spherical triangle with vertex C at the North Pole and vertices A and B at a geocentric latitude of 45[degrees].
Hence it is evident, that every visible right-lined triangle, will coincide in all its parts with some spherical triangle.
Every visible triangle is visibly indistinguishable from some spherical triangle.
Suppose we take literally (as I initially refused to do) Reid's claim in paragraph 5 that every visible triangle coincides in all its parts with some spherical triangle.
Every visible triangle coincides in all its parts with, and is thus identical with, some spherical triangle.
The angle from north to the velocity vector can be determined using the right spherical triangle formed by the velocity-vector azimuth and elevation angles.
Then, from the spherical triangle between N, V and R, using the law of cosines for sides, we find the angle between the transmitter's velocity vector and the receiver:
The two right spherical triangles and one spherical triangle of Fig 5 allow the calculation of the angular distance from the roll axis and the target (j):
AREAS Real function which returns the area of a spherical triangle on the unit sphere.
CIRCUM Subroutine which computes the circumcenter of a spherical triangle defined by user-specified vertices on the unit sphere.
The "sides" of the spherical triangle must be great circles of the unit sphere -- that is, they must be the intersection of the surface of the sphere with a plane passing through the origin of the sphere.
For example, all three of the "angles" in a spherical triangle could be 90[degree].
Trigonometric Relationships in Any Spherical Triangle