Phi vs Theta referring to angles of vectors Even though this question arises from doing simple vector math from a physics textbook I figured it's much more general than that application.
The problem is as follows:
A spelunker is surveying a cave. She follows a passage 180m straight west, then 210m in a direction 45 degrees east of south, and then 280m at 30 degrees east of north. After a fourth unmeasured displacement she finds herself back where she started. Determine the magnitude and direction of the fourth displacement.
My work is as follows:
Let the magnitude of $\vec{A}$ be 180m. Let the magnitude of $\vec{B}$ be 210m and it's direction be 45 degrees east of south.  Let $\vec{C}$ have the magnitude 280m and a direction of 30 degrees east of north.
$$0 = \vec{A} + \vec{B} + \vec{C} + \vec{D}$$
$$\vec{D} = -(\vec{A} + \vec{B} + \vec{C})$$
$$D_x = -(-180 + 210\sin{45} + 280\sin{30}) = -108.5m$$
$$D_y = -(0 - 210\cos{45} + 280\cos{30}) = -94m$$
$$|D| = \sqrt{(-108.5)^2 + (-94)^2} = 144m$$
$$D_\theta = \arctan{D_y/D_x} = \arctan{-94/-108.5} = 41 \deg$$
Looking over the answers for the angle of a result vector that sits in quadrant 3 I get two possible answers according to the book (the result of the $\arctan{Ry/Rx} = 41$):
$\theta = 180 + 40.9$ and $\Phi = \theta - 180$
The way I approached the problem was to recognize that (in the case of this problem) the x and y component of the resultant were in quadrant 3 and then I just drew an angle 41 degrees to the south of the -x axis. My math corresponded to the $\Phi$ interpretation. However, I don't understand why there are two different definitions of the resultant angle. I understand they're both basically the same (theta sets the angle in Q1 and rotates it whereas phi is simply the angle) but I can't figure out why you need mention both then. Can anyone help me? 
 A: It is a common mistake for people to confuse

The arctan of $x$

with

The solution to $\tan \theta = x$

For every value of $x$, the equation has infintiely many solutions, and it has two different solutions with $\theta \in [0^\circ, 360^\circ)$ (or in any interval of length $360^\circ$ you choose). Similar statements are true for each of the trig functions.
We can see the dangers of making this mistake in your work: although you compute a vector $\vec{D}$ in the third quadrant, the polar form you computed is for a vector in the first quadrant!
The $\tan$ function satisfies $\tan \theta = \tan(\theta + n 180^\circ)$ (where $n$ can be any integer). Within the interval $[0^\circ, 360^\circ)$, the other solution that you missed is thus $41^\circ + 180^\circ$, which does give an angle in the third quadrant.

Some methods of calculation provide an alternative to $\arctan$ to avoid this issue; rather than computing an angle via $\arctan(y/x)$, they offer a different version of $\arctan$ that takes two arguments, and always gives an angle in the correct quadrant for the vector $(x,y)$. e.g.


*

*In WolframAlpha we can compute arctan(x,y).

*In python, we can compute math.atan2(y,x).

