# Finding the Direction when resolving a Vector

When you are finding the resultant of a vector I understand that to find the magnitude you use Pythagorean Theorem and I understand that to find the direction you are going to use $\tan^{-1}$ (or corresponding trig function). My question is how do you decide what angle you are finding? There are two angles and is there a convention for which one you solve for?

I was also wondering, why do you have to add 90 or 180 degrees to vectors outside of the first quadrant? Can you not give the angle and direction?

For example, with the problem:

Find the resultant of the vectors 10 km North and 5 km West. The magnitude is 11.18 km, but exactly how would you decide the angle for direction and why would you have to add 90 to it? (The answer is 26.6 degrees + 90= 116.6 degrees) Could I not say that the resultant is 11.18 km, 26.6 degrees NorthWest? And why do you chose that angle to work with, moving the 5 km vector parallel to complete the triangle instead of the 10 km vector?

Edit: In regards to the sketch below, why is the one on the right considered correct and why would I have to add 90 to it? There is no hard and fast convention. What is important is that the direction you state is unambiguous. You might state a given angle as $20^\circ$ South of West, or $200^\circ$ counterclockwise from East. Neither is "more correct", but note that neither is ambiguous, either.

Typically in motion problems angles tend to be measured from the horizontal. For example, in projectile motion problems, the launch angle of the projectile is nearly always measured from the horizontal, and when doing force inclined-plane problems, the same is usually true. However, there are situations where that is not always true. Sometimes in statics problems angles are measured from the vertical.

Additionally, although a little far afield from what you're talking about, in optics, the angles of incident light rays are conventionally measured from the normal.

However, no matter how you are measuring the direction of your vector, it's good practice to be as explicit as possible so as not to confuse anyone. An answer like "$\theta$ degrees North of East" is pretty good. Alternately, you could say "$\theta$ degrees above (or below) the horizontal" as long as you still specify the left or right direction as well. "A cannonball is shot to the right, at an angle of 30 degrees above the horizontal" would be an example of this way. Note that the direction is still fully specified.

There is a mathematical convention for direction angles $\alpha$ in an $x$-$y$ diagram: counter-clockwise from positive horizontal axis (your "east" direction). Since $\tan \alpha = y/x$ is a periodic function with period $\pi$, $\arctan (y/x)$ has two solutions in the range $[0, 2\pi]$, while your calculator may give you just one solution, mostly in the range $[-\pi,\pi]$ or $[-90°,90°]$, respectively.

To find the "right" solution, you have to consider the signs of your $x$ and $y$ values. "North" means positive $y$, "west" means negative $x$ value. Therefore your direction vector is in the second (northwest) quadrant, i.e. an angle in the range $[\pi/2,\pi]$ or $[90°,180°]$.

P.S. The solution does not depend on whether to go to north first and after that to the west or vice versa, like your diagramm suggests. The ambiguity stems from the division $y/x$. So you can have the same line slope by moving in two opposite directions ("north-west" or "south-east").