Question

How do you find the angle `alpha` such that the angle lies in quadrant III and `secalpha=-1.108`?

Answer 1

`205^@84`

Explanation:

`sec a = 1/(cos a) = - 1.108`
`cos a = - 1/1.108 = - 0.90`
cos a = - 0.90 -->calculator gives: arc `a = 154^@16`
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.