Algebra 2 (1st Edition)

$260.54^{\circ}$
Here, we have $\sin^{-1} (6)$ The angle $\theta$ is in the First quadrant. This means that $\theta=\tan^{-1} (6) \approx 80.54^{\circ}$ We want an angle in the Third quadrant: Then we have $\theta=\theta'+180^{\circ}=80.54^{\circ}+180^{\circ}=260.54^{\circ}$