## Algebra 2 (1st Edition)

$-\dfrac{\sqrt 3}{2}$
Here, we have $\cos 150^{\circ}$ This angle lies in the Second Quadrant. Here, $\theta'=\pi-\theta=180^{\circ}-150^{\circ}=30^{\circ}$ Now, $\cos \theta'=\cos 30^{\circ}=\dfrac{\sqrt 3}{2}$ The function cosine is negative in Quadrant $II$. Thus, $\cos \theta= -\cos \theta'=-\dfrac{\sqrt 3}{2}$