Question

If `sec theta = 5/3` and the terminal point determined by `theta` is in quadrant 4, what is the value of `theta`?

Answer 1

`t = - 53^@13`

Explanation:

`sec t = 5/3` --> `cos t = 3/5`
Calculator and unit circle give 2 solutions:
`t = +- 53^@13`
Since t is in Quadrant 4, t is negative -->
`t = - 53^@13`