Answer to Question #342730 in Trigonometry for Ton

Question

Solve the spherical triangle right angled C given a= 23° 10' and c = 25° 12'

Answer 1

In Spherical Trigonometry, there exist the following Law of Cosines and Law of Sines, for vertices and sides of triangles… which are all angles.

(I) cos(a) = cos(b)cos(c) + sin(b)sin(c)cos(A)

(II) sinA/sin(a) = sinB/sin(b) = sinC/sin(c)

For side ‘b’, we have, using (I): cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(C) = cos(a)cos(b) since C = 90º

==> cos(b) = cos(c)/cos(a) = cos(25º12')/cos(23º10') ==> b = 10.26º

For vertex angles A, B, use (II):

sinC/sin(25º12') = sinA/sin(23º10') = sinB/sin(b)

sinA = sin(23º10')/sin(25º12') ==> A = 67.52º

sinB = sin(10.26º) / sin(25º12') ==> B = 24.7º