Given Point A and Point B in 2D space, how can I find the angle Point B is from Point A? 0° can be any direction; it doesn't matter. For example, Point A is at (0, 10) and Point B is at (10, 20). The angle is 45° in this example (assuming 0° is up).
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I believe that the accepted answer does not correctly solve the problem for many people visiting this question. I am doing 2D simulations and the answer above does not solve my problem. Imagine the following circle:
$\hskip1.8in$
Assume that the middle of the circle is point A. Point B is at some angle from A according to the angles of the circle (so 0°) is right.
$\hskip2in$The atan2 function is what you need!
$$ atan2(y, x) $$ $\hskip3.2in$ Where
$$ y = y_B - y_A $$ $$ x = x_B - x_A $$
Thomas Wagenaar
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2You made my day. I spent hours trying to figure this out and got nothing but convoluted solutions that never return a proper angle. Thank you! – KingOfHypocrites Aug 28 '18 at 17:19
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Is x1,y1 point A, and x2,y2 point B? – Steve Smith May 16 '20 at 15:21
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@SteveSmith yes. – Thomas Wagenaar Jul 02 '20 at 12:21
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Maybe this is useful to someone: https://justingolden.me/angleto/ – Justin Mar 27 '21 at 02:41
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1Note that some implementations are ${\rm atan2}(\Delta x,,\Delta y)$ and some ${\rm atan2}(\Delta y,,\Delta x)$. Check your documentation. – John Alexiou Nov 07 '22 at 20:14
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From what I understood about your question, you want to find the angle between two points given their coordinates. In that case, first find the slope of the line using the slope formula: $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are coordinates on the line. Next, use this formula: $$\tan(\theta)=m$$ where $\theta$ is the angle. Therefore, the angle $\theta$ equals: $$\theta=\tan^{-1}(m)$$
Let's use the points $(0,10)$ and $(10,20)$ as an example (you mentioned it in your question). The slope is: $$m=\dfrac{10-20}{0-10}$$ $$m=\dfrac{-10}{-10}$$ $$m=1$$ Now we will find $\theta$. $$\tan(\theta)=1$$ $$\theta=\tan^{-1}(1)$$ $$\theta=45^\circ$$
Note: The $\tan(\theta)=m$ formula only gives the angle facing the positive $x$-axis (i.e. facing the "right"). So for a negative slope, you should get an angle that is greater than $90^\circ$.
Anonymous Computer
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In your example using my two points you used the Xs over the Ys. Is that intentional? Why does it differ from the equation above? – Keavon Mar 14 '14 at 02:08
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@Keavon Nice observation, it is supposed to be the Ys over the Xs, sorry – Anonymous Computer Mar 14 '14 at 03:19
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1What about the scenario where A = (10, 0) and B = (10, 10), and so x2-y2 = 0 and you get a divide by 0 error? This isn't a paticularly helpful answer imo. – user1068446 Jul 15 '18 at 02:42
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