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

$- {22}^{\circ} 88$
$\sin t = \frac{1}{\csc} = - \frac{1}{2.572} = - 0.388$
sin t = - 0.388 --> arc $t = - {22}^{\circ} 88$