Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 15

$136.05^{\circ}$

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}$