Question

How do you find the angle `alpha` such that the angle lies in quadrant IV and `cscalpha=-2.572`?

Answer 1

`- 22^@88`

Explanation:

csc t = -2.572
`sin t = 1/(csc) = - 1/2.572 = - 0.388`
Use calculator:
sin t = - 0.388 --> arc `t = - 22^@88`