# 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?

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Without the problem that gave rise to your question, it's hard to answer. Consider supplying the problem and brief work, and then we can frame the answer for you. –  rschwieb Feb 23 '13 at 13:17
Supplied work for you. –  anonacct1405 Feb 23 '13 at 14:22

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.

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So I suppose then I must infer that with Dx and Dy indicating the third quadrant I should be finding the angle and adding or subtracting the proper degrees to place the direction of the vector in the correct place? How do they obtain phi as the right answer as well then (the book says either 41 or 221 degrees are correct)? Thank you! –  anonacct1405 Feb 23 '13 at 18:51
Right. As for the final answer, there are a number of ways to specify directions. An angle of 221 degrees (measured counterclockwise from East) would most commonly be stated as "41 degrees South of West" when stating a direction in "real life". This is especially important in "real life" because there's another convention that measures angles clockwise from North. So "221 degrees" is actually ambiguous, and could mean "41 degrees South of West" or "41 degrees West of South" depending on which convention one uses. –  Hurkyl Feb 23 '13 at 23:14
So phi would simply represent the reference degrees rather than the semi-ambiguous terminal angle method. Got it. Thanks again! –  anonacct1405 Feb 24 '13 at 0:39