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So when I am working on homework and I am asked to find the direction angle of a vector I usually to the inverse of tan(y/x). Whatever answer I get, I usually add or subtract from the answer 180 or 360 to get the direction angle. However, I do not know which one to do in which situation.

For example, on one of my problems it was <-10, 9>. I took the inverse of tan(9/-10) and added 360° to the result for an answer of 138.01°.

Another example, <-6,0>,I took the inverse of tan(0/-6) and added 180° for an answer of 0°.

I'm just not sure when exactly to do which. I got the answer correct because I had an answer sheet to check my work, but if I didn't get the answer right the first time I would try something else. That won't work on my exam when I don't have an answer sheet to check my homework.

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  • $\begingroup$ The signs of the components of the vector tell you what quadrant it’s in, which narrows the angle down to a 90° range. $\endgroup$ – amd Apr 19 '17 at 20:22
  • $\begingroup$ Most calculators or programming environments have a function named atan2(dy,dx) that find the correct angle. $\endgroup$ – ja72 Apr 20 '17 at 19:37
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Draw a picture beforehand and you will have some kind of idea where your angle should lie. In particular, if you know your unit circle very well, you will know what angles correspond with which quadrants.

So for the vector $\langle-10,9\rangle$, we know the $x$-component is negative meaning it goes to the left, and its $y$-component is positive, meaning it moves up. So on a coordinate plane, you know that this ends up in quadrant II. In quadrant II, you deal with angles between $90^\circ$ and $180^\circ$. So, the answer for $\approx 138^\circ$ is reasonable to leave as-is.

When you're dealing with $\langle-6,0\rangle$, if we draw a picture, the $x$-component makes the vector go left, and the $y$-component contributes nothing to the direction of the vector. So if we take the positive $x$-axis to be $0^\circ$, then the negative $x$-axis will be $180^\circ$. Hence, it is obvious that $\tan^{-1}\left({0\over-6}\right) = 0^\circ$ is not reasonable to leave as-is, and why we must add $180^\circ$ to the angle measure.

Let's try one more example, shall we? Consider the vector $\langle-3,-4\rangle$. This vector ends up in quadrant III, which means the angle should end up between $180^\circ$ and $270^\circ$. Then $\tan^{-1}\left({-4\over-3}\right) \approx 53^\circ$, which is in quadrant I. However, this does not line up with our intuition of where this vector should land. Hence, we may add $180^\circ$ to this degree measure to give us the correct angle measure which will land in quadrant III, since $53^\circ + 180^\circ = 233^\circ$.

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    $\begingroup$ This is an excellent answer, +1. The only thing you didn't explain was WHY one needs to add 180º or 360º, i.e., the range of arctan is (-90º, 90º). $\endgroup$ – scott Apr 19 '17 at 20:38
  • $\begingroup$ @scott, that is a very important observation to make. Thank you. $\endgroup$ – Decaf-Math Apr 20 '17 at 18:57
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A formula that is giving the angle in degrees .

f(x,y)=180-90*(1+sign(x))* (1-sign(y^2))-45*(2+sign(x))*sign(y)

  -180/pi()*sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))
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  • $\begingroup$ Please, use Mathjax (i.e. LaTeX commands) for mathematical notations. $\endgroup$ – Taroccoesbrocco Nov 18 '18 at 6:41

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