Chapter 12 - Section 12.2 - Vectors - 12.2 Exercises - Page 844: 29

$v=⟨-2\sqrt 3, 2⟩$

We know that $|v|=4$ and $\theta=\frac{5\pi}{6}$ where $\theta$ is the angle formed with the positive $x$ axis. We can compute the components of $|v|$ using the trigonometric functions as follows: $v_x=|v|\cos{\theta}=4 \cdot \cos{\frac{5\pi}{6}}=-2\sqrt3$ $v_y=|v|\sin{\theta}=4 \cdot \sin{\frac{5\pi}{6}}=2$ $$v=⟨-2\sqrt3,2⟩$$