Why am I getting this vector question wrong about finding the angle? EDIT: this question is about how to use trig to find the angle between the x-axis and a vector. Just editing since stack exchange asked me to clarify what concept I am asking about.
This is a first year physics problem, I'm studying for my exam and am doing some practice and a bit of help would be appreciated for this question. My answer is very close to the answer key but still wrong, and I want to know how and why.
Here is the problem

If the velocity vector of a polar bear is ($\vec u = -18.0 \hat i - 13.0\hat j$)
, how fast and in what
geographic direction is it heading? Here, $\hat i$ and $\hat j$ are directions to geographic east and north, respectively

What I've gotten so far is I've used the cosine law $\vec u 
 \cdot \hat i = |\vec u||\hat i|cos\theta$ to find that
$-18 = (\sqrt{493})(1)cos\theta $
$-0.8106792284 = cos\theta $
$\theta = cos^{-1}(-0.8106792284) $
$\theta = 144.162347$
which I work out to be $35.8^{\circ}$ north of west, but the answer is $35.8^{\circ}$ south of west.
My error probably has something to do with the fundamentals of trig but just for context, I'm studying for a deferred exam so I haven't touched any math or physics material for about 18 months!
 A: The answer is quite simple:
$$\cos(\theta)=-0.81067...$$
so one of the solutions is:
$$\theta = \cos^{-1}(-0.81067) = 144.16 \deg$$
but remember:
$$\cos(\theta)=\cos(-\theta)$$
so both $\theta$ and $-\theta$ give the same answer (similar to how $x^2=1$ has two solutions. So the other solution is:
$$\theta = -\cos^{-1}(-0.81067) = -144.16 \deg$$
Which is what you're looking for. How do you know which one to choose? well since both $i$ and $j$ are negative, then you know it has to be somewhere south-west, which is the second solution.
A: $\cos^{-1}(-x) = \pi-cos^{-1}{x}$
This answer applies when you are trying to find the solution in the principal branch. You could have checked the sign by drawing the vector in the cartesian coordinate system.
I use the formula $\theta = \tan^{-1}{\dfrac{y}{x}}$. We know that both sines and cosines are negative in the third quadrant; hence, the angle should also lie between $\pi$ and $\dfrac{3\pi}{2}$.
I suggest not relying on calculators for these calculations as the answers you get are only in the principal branch.
