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How to calculate an angle from three points?

Asked 2009-07-31T07:55:04.800
120

Lets say you have this:

P1 = (x=2, y=50)
P2 = (x=9, y=40)
P3 = (x=5, y=20)

Assume that P1 is the center point of a circle. It is always the same. I want the angle that is made up by P2 and P3, or in other words the angle that is next to P1. The inner angle to be precise. It will always be an acute angle, so less than -90 degrees.

I thought: Man, that's simple geometry math. But I have looked for a formula for around 6 hours now, and only find people talking about complicated NASA stuff like arccos and vector scalar product stuff. My head feels like it's in a fridge.

Some math gurus here that think this is a simple problem? I don't think the programming language matters here, but for those who think it does: java and objective-c. I need it for both, but haven't tagged it for these.

algorithm math geometry
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2 Answers

47

It gets very simple if you think it as two vectors, one from point P1 to P2 and one from P1 to P3

so:
a = (p1.x - p2.x, p1.y - p2.y)
b = (p1.x - p3.x, p1.y - p3.y)

You can then invert the dot product formula:
dot product
to get the angle:
angle between two vectors

Remember that dot product just means: a1*b1 + a2*b2 (just 2 dimensions here...)

answered 2009-08-30T15:56:05.117
12

If you are thinking of P1 as the center of a circle, you are thinking too complicated. You have a simple triangle, so your problem is solveable with the law of cosines. No need for any polar coordinate tranformation or somesuch. Say the distances are P1-P2 = A, P2-P3 = B and P3-P1 = C:

Angle = arccos ( (B^2-A^2-C^2) / 2AC )

All you need to do is calculate the length of the distances A, B and C. Those are easily available from the x- and y-coordinates of your points and Pythagoras' theorem

Length = sqrt( (X2-X1)^2 + (Y2-Y1)^2 )

answered 2009-07-31T08:08:00.077

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