How do you find the angle #alpha# such that the angle lies in quadrant III and #secalpha=-1.108#?
cos a = - 0.90 -->calculator gives: arc
Unit circle gives another arc x that has the same cos value:
x = - 154^@16 or
x = 360 - 154.16 = 205^@84.
This arc lies in Quadrant 3.