Answer
$\lt 4, 6.93 \gt $
Work Step by Step
Let $v$ is a vector that has a magnitude $|v|$ and the direction angle of $\theta$.
The magnitude of horizontal component is given by: $v_x=|v| \cos \theta= 8 \cos 60^{\circ}=4$
And
The magnitude of vertical component is given by: $v_y=|v| \sin \theta=8 \sin 60^{\circ}= 6.93$
Thus, $v=\lt 4, 6.93 \gt $