Answer
$\lt -1.53, -129 \gt $
Work Step by Step
Let $v$ is a vector that has a magnitude $|v|$ and the direction angle of $\theta$.
Firstly, we need to convert the given angle to standard position.
That is, $ \theta =360^{\circ}- 140^{\circ} = 220^{\circ}$
The magnitude of horizontal component is given by: $v_x=|v| \cos \theta= 2 \cos 220^{\circ}= -1.53$
And
The magnitude of vertical component is given by: $v_y=|v| \sin \theta= 2 \sin 220^{\circ}= -1.29$
Thus, $v=\lt -1.53, -129 \gt $
