## Algebra 2 (1st Edition)

$136.05^{\circ}$
Here, we have $\cos \theta =-0.72; 90^{\circ} \lt \theta \lt 180^{\circ}$ $\theta'=\cos^{-1} (0.72) \approx 136.05^{\circ}$ This angle $\theta'$ is in the $II$ quadrant. Then we have $\cos 136.05^{\circ} \approx -0.719945 \approx -0.72^{\circ}$