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I don't understand why this is incorrect. It's y/x times tan inverse, right?

enter image description here

iadvd
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user416503
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    You are close, but in the wrong quadrant. Consider the signs of $x,y,$. – dxiv Feb 20 '17 at 04:07
  • Try to consider $\tan\theta=\frac{4.49}{-4.15}$. So $\theta$ is in the $II$-quadrant. – Juniven Acapulco Feb 20 '17 at 04:13
  • I don't understand how it can be in Q2. I sketched the graph and it should be in Q4, just like my answer said - http://i.imgur.com/KBUYJzp.jpg – user416503 Feb 20 '17 at 04:20
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    In what quadrant does the point $(-4.15,4.49)$ belongs? – Juniven Acapulco Feb 20 '17 at 04:24
  • it's just like how i sketched in the above pic... neg x and pos y is in Q4. – user416503 Feb 20 '17 at 04:28
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    @user416503 Negative $x$ is to the left of the $y$ axis, so it can't possibly be in Q4. – dxiv Feb 20 '17 at 04:30
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    Remember that $arctan$ and $arcsin$ always give a result in quadrants $I$ and $IV$. If the vector is actually in one of the other two quadrants you have to add $180^\circ$ or $\pi$ (if you are using radians). – John Wayland Bales Feb 20 '17 at 04:41
  • Notwithstanding quadrant issues, what do you mean by times tan inverse? I suggest you learn the general concept of functions and function notation as soon as possible. – zipirovich Feb 20 '17 at 04:50

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