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

Feb 18, 2017

${205}^{\circ} 84$

#### Explanation:

$\sec a = \frac{1}{\cos a} = - 1.108$
$\cos a = - \frac{1}{1.108} = - 0.90$
cos a = - 0.90 -->calculator gives: arc $a = {154}^{\circ} 16$
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.