If r = sqrt(3)i + j, find the angle that theta makes with x axis.

The given vector is $r = \sqrt{3}i + j$.
Here, the $x$-component is $x = \sqrt{3}$ and the $y$-component is $y = 1$.

The angle $\theta$ that a vector $(x, y)$ makes with the positive $x$-axis is given by the formula:

Substitute the values of $x$ and $y$:

We need to find the angle whose tangent is $\frac{1}{\sqrt{3}}$.
We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Since both $x$ and $y$ components are positive, the vector lies in the first quadrant, so the angle is $30^\circ$.

The angle is $30^\circ$. Comparing this with the given options, option A is $30^\circ$.

The final answer is $\text{A. } 30^\circ$.