Answer
$\theta = 60 ^{\circ}$ (or $\frac{\pi}{3})$.
Work Step by Step
The given vector has x-component 1 and y-component $\sqrt 3$. Thus, this vector can be drawn as an arrow going from the origin to the point (1, $\sqrt 3$). The vector can then be considered the hypotenuse of a right triangle with horizontal leg 1 and vertical leg $\sqrt 3$. If we let $\theta $ be the angle between the positive direction of the x-axis and the vector, then we can write:
$\tan \theta = \frac {\sqrt 3}{1}$
To get $\theta$, we can use the inverse tangent function, or we can recognize that in the first quadrant, this is the tangent of $\theta = 60 ^{\circ}$ (or $\frac{\pi}{3})$.
