Answer to Question #342728 in Trigonometry for Ton

Question

Solve the spherical triangle ABC right angled C given c= 56° 13' and b = 48°30'

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

==> cos(a) = cos(c)/cos(b) = cos(56º13')/cos(48º30') ==> b = 32.97º

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

sinC/sin(56°13') = sinA/sin(48º30') = sinB/sin(b)

sinA = sin(48º30')/sin(56º13') ==> A = 64.16º

sinB = sin(32.97º) / sin(56º13') ==> B = 40.92º