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.
If you mean the angle that P1 is the vertex of then using the Law of Cosines should work:
arccos((P122 + P132 - P232) / (2 * P12 * P13))
where P12 is the length of the segment from P1 to P2, calculated by
sqrt((P1x - P2x)2 + (P1y - P2y)2)