How do I find the angle a vector makes to the $+x$ axis? You are given two vectors $\vec a = -3.00\hat i + 7.00\hat j$ and $\vec b= 4.00\hat i + 2.00\hat j$. Let the counterclockwise angles be positive.

What angle $\theta (\vec a)$ where $0^\circ \le \theta (\vec a) &lt; 360^\circ $, does $\vec a$ make with the $+x$-axis?

I drew a right triangle with a $\vec ax$ component of $-3$ and an $\vec ay$ component of $7$. Do I just use trig to find the angle off the $x$-axis?
A general rule for finding the counterclockwise angle from the positive $x$ axis to the direction of a two-dimensional vector is explained in
How to convert components into an angle directly (for vectors)?
The basic idea is that you first take the arc tangent of the $y$-component
divided by the $x$-component:
$$
\theta_1 = \arctan\frac{a_y}{a_x}.
$$
One complication is that this procedure gives the same resulting $\theta_1$
for the vector $\langle a_x,a_y \rangle$ and the vector $\langle -a_x,-a_y \rangle,$ which are two vectors in directions $180$ degrees apart.
So $\theta_1$ may be in the direction you want, or in the direction
$180$ degrees opposite.
In fact $\theta_1$ will be in the correct direction whenever $a_x &gt; 0,$
since that is how the arc tangent function is designed to work.
(It always produces angles in the range $-\frac\pi2$ to $\frac\pi2$ radians,
that is, $-90$ to $90$ degrees,
which correspond to vectors with positive $x$-components.)
The cases where $\theta_1$ is wrong are precisely the cases where
$a_x &lt; 0,$ so if $a_x &lt; 0$ (as it is in your particular question)
you have to add $180$ degrees to $\theta_1.$
A second complication (which does not actually come up in your particular
question) occurs because the procedures given above produce a result in the range $-90$ degrees to $270$ degrees, but people usually want an answer in the range $-180$ to $180$ or $0$ to $360.$ The solution to this, of course, is to add or subtract $360$ degrees as needed to get an answer in the desired angle range.
Yeah you just use trig. Tan is the easiest. I saw your question about whether it is arctan(7/3) or arctan(7/-3) it is the 7/-3. Now this will give you a negative angle. This is outside your range so add 360 to it to get the answer that is in your range.