## University Calculus: Early Transcendentals (3rd Edition)

$\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$
Let the components of vector $v$ be given as $v=\lt v_x,v_y\gt$ Now, $v_x=1 \cos (\dfrac{\pi}{6})=\dfrac{\sqrt 3}{2}$ and $v_y=1 \sin (\dfrac{\pi}{6})=\dfrac{1}{2}$ Hence, $v=\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$