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

Question

1488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T04:30:02

$- 22^@88$

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

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