Dtyjlui

You are given two vectors $vec a = -3.00hat i + 7.00hat j$ and $vec b= 4.00hat i + 2.00hat j$. Let the counterclockwise angles be positive.

What angle $theta (vec a)$ where $0^circ le theta (vec a) < 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?

You can use trig, but since these are vectors, there's a far easier way. Use dot products!

Suppose you have a vector $vec{a}$ and you need to find the angle it makes with the $x$-axis. So take a unit vector along the $x$-axis, viz. $hat{i}$. Using the dot product of these vectors,

$$
vec{a}cdothat{i} = |vec{a}||hat{i}|costheta
$$

where $theta$ is the angle between the two vectors. Since $|hat{i}|$ = 1, and $vec{a}cdothat{i} = a_x$, hence

$$
costheta = frac{a_x}{|vec{a}|} implies theta = cos^{-1} frac{a_x}{|vec{a}|}.
$$

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:
$$
theta_1 = arctanfrac{a_y}{a_x}.
$$

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 $a_x > 0,$
since that is how the arc tangent function is designed to work.
(It always produces angles in the range $-fracpi2$ to $fracpi2$ 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
$a_x < 0,$ so if $a_x > 0$ (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.

Problem: Counterclockwise angle between $ +x-$ axis and vector $vec{a} = (-3,7)$.

Draw a $ x-,y- $ axes, and vector $vec{a}$ pointing from the origin $(0,0) $ to $ (-3,7)$. Verify that we are in the second quadrant , I.e. $angle theta $ is obtuse.

Normalize vector $vec{a}$ , call it

$vec{n}$ : $(1/√58) (-3,7)$.

Dot product: $vec{n} cdot vec{e_x}$ = $cos(theta)$.

$cos(theta)$ = $(1/√58) (-3, 7) cdot (1, 0) $= $-3 (1/√58)$.

Left: Look up $arccos ( theta)$ to find the obtuse angle $theta$.

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.