Answer
See below
Work Step by Step
Using the given triangle, we find:
$$ \cos \theta_1= \frac{2\sqrt 3}{4} \\ \theta_1 = \arccos^{-1}(\frac{\sqrt 3}{2}) \\ \theta_1 = 30^{\circ}$$
Since $\theta=\theta_1+90^{\circ}\\\theta=30^{\circ}+90^{\circ}\\\theta=120^{\circ}$
