How to calculate an angle from three points

Question

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

User · Answer

Here s a C  method to return the angle  0-360  anticlockwise from the horizontal for a point on a circle       public static double GetAngle Point centre  Point point1                   Thanks to Dave Hill            Turn into a vector  from the origin          double x   point1 X - centre X          double y   point1 Y - centre Y             Dot product u dot v   mag u   mag v   cos theta            Therefore theta   cos -1   u dot v     mag u   mag v              Horizontal v    1  0             therefore theta   cos -1  u x   mag u             nb  there are 2 possible angles and if u y is positive then angle is in first quadrant  negative then second quadrant         double magnitude   Math Sqrt x   x   y   y           double angle   0          if magnitude  gt  0              angle   Math Acos x   magnitude            angle   angle   180   Math PI          if  y  lt  0              angle   360 - angle           return angle          Cheers  Paul

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The best way to deal with angle computation is to use atan2 y  x  that given a point x  y returns the angle from that point and the X  axis in respect to the origin   Given that the computation is  double result   atan2 P3 y - P1 y  P3 x - P1 x  -                 atan2 P2 y - P1 y  P2 x - P1 x     i e  you basically translate the two points by -P1  in other words you translate everything so that P1 ends up in the origin  and then you consider the difference of the absolute angles of P3 and of P2   The advantages of atan2 is that the full circle is represented  you can get any number between -p and p  where instead with acos you need to handle several cases depending on the signs to compute the correct result   The only singular point for atan2 is  0  0     meaning that both P2 and P3 must be different from P1 as in that case doesn t make sense to talk about an angle

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Recently  I too have the same problem    In Delphi It s very similar to Objective-C   procedure TForm1 FormPaint Sender  TObject   var ARect  TRect      AWidth  AHeight  Integer      ABasePoint  TPoint      AAngle  Extended  begin   FCenter    Point Width div 2  Height div 2     AWidth    Width div 4    AHeight    Height div 4    ABasePoint    Point FCenter X AWidth  FCenter Y     ARect    Rect Point FCenter X - AWidth  FCenter Y - AHeight       Point FCenter X   AWidth  FCenter Y   AHeight      AAngle    ArcTan2 ClickPoint Y-Center Y  ClickPoint X-Center X    180   pi    AngleLabel Caption    Format  Angle is  5 2f    AAngle      Canvas Ellipse ARect     Canvas MoveTo FCenter X  FCenter Y     Canvas LineTo FClickPoint X  FClickPoint Y     Canvas MoveTo FCenter X  FCenter Y     Canvas LineTo ABasePoint X  ABasePoint Y   end

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Let me give an example in JavaScript  I ve fought a lot with that          Calculates the angle  in radians  between two vectors pointing outward from one center        param p0 first point     param p1 second point     param c center point     function find angle p0 p1 c        var p0c   Math sqrt Math pow c x-p0 x 2                           Math pow c y-p0 y 2       p0- gt c  b         var p1c   Math sqrt Math pow c x-p1 x 2                           Math pow c y-p1 y 2       p1- gt c  a      var p0p1   Math sqrt Math pow p1 x-p0 x 2                            Math pow p1 y-p0 y 2       p0- gt p1  c      return Math acos  p1c p1c p0c p0c-p0p1 p0p1   2 p1c p0c        Bonus  Example with HTML5-canvas

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well  the other answers seem to cover everything required  so I would like to just add this if you are using JMonkeyEngine   Vector3f angleBetween otherVector   as that is what I came here looking for

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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   to get the angle    Remember that  just means  a1 b1   a2 b2  just 2 dimensions here

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there IS a simple answer for this using high school math    Let say that you have 3 points  To get angle from point A to B  angle   atan2 A x - B x  B y - A y   To get angle from point B to C  angle2   atan2 B x - C x  C y - B y   Answer   180   angle2 - angle If  answer  lt  0       return answer   360  else      return answer     I just used this code in the recent project that I made  change the B to P1   you might as well remove the  180    if you want

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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

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In  Objective-C you could do this by  float xpoint      atan2  newPoint x - oldPoint x     newPoint y - oldPoint y    180  M PI     Or read more here

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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

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Very Simple Geometric Solution with Explanation  Few days ago  a fell into the same problem  amp  had to sit with the math book  I solved the problem by combining and simplifying some basic formulas     Lets consider this figure-    We want to know    so we need to find out a and    first  Now  for any straight line-   y   m   x   c   Let- A    ax  ay   B    bx  by   and O    ox  oy   So for the line OA-  oy   m1   ox   c     c   oy - m1   ox       eqn-1   ay   m1   ax   c     ay   m1   ax   oy - m1   ox    from eqn-1                       ay   m1   ax   oy - m1   ox                      m1    ay - oy     ax - ox                       tan a    ay - oy     ax - ox     m   slope   tan          eqn-2    In the same way  for line OB-  tan       by - oy     bx - ox        eqn-3    Now  we need        - a  In trigonometry we have a formula-  tan    -a     tan      tan a     1 - tan      tan a        eqn-4    After replacing the value of tan a  from eqn-2  and tan b  from eqn-3  in eqn-4  and applying simplification we get-  tan    -a       ax-ox   by-oy   ay-oy   bx-ox         ax-ox   bx-ox - ay-oy   by-oy      So         -a   tan  -1      ax-ox   by-oy   ay-oy   bx-ox       ax-ox   bx-ox - ay-oy   by-oy       That is it      Now  take following figure-    This C  or  Java method calculates the angle    -      private double calculateAngle double P1X  double P1Y  double P2X  double P2Y              double P3X  double P3Y            double numerator   P2Y  P1X-P3X    P1Y  P3X-P2X    P3Y  P2X-P1X           double denominator    P2X-P1X   P1X-P3X     P2Y-P1Y   P1Y-P3Y           double ratio   numerator denominator           double angleRad   Math Atan ratio           double angleDeg    angleRad 180  Math PI           if angleDeg lt 0               angleDeg   180 angleDeg                     return angleDeg

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I ran into a similar problem recently  only I needed to differentiate between a positive and negative angles  In case this is of use to anyone  I recommend the code snippet I grabbed from this mailing list about detecting rotation over a touch event for Android     Override  public boolean onTouchEvent MotionEvent e        float x   e getX        float y   e getY        switch  e getAction          case MotionEvent ACTION MOVE           find an approximate angle between them          float dx   x-cx         float dy   y-cy         double a Math atan2 dy dx           float dpx  mPreviousX-cx         float dpy  mPreviousY-cy         double b Math atan2 dpy  dpx           double diff    a-b         this bearing -  Math toDegrees diff          this invalidate              mPreviousX   x      mPreviousY   y      return true

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Atan2        output in degrees        PI 2               90                                                                 PI --- --- 0    180 --- --- 0                                                                       -PI 2              270  public static double CalculateAngleFromHorizontal double startX  double startY  double endX  double endY        var atan   Math Atan2 endY - startY  endX - startX      Angle in radians     var angleDegrees   atan    180   Math PI       Angle in degrees  can be   -      if  angleDegrees  lt  0 0                angleDegrees   360 0   angleDegrees            return angleDegrees        Angle from point2 to point 3 counter clockwise public static double CalculateAngle0To360 double centerX  double centerY  double x2  double y2  double x3  double y3        var angle2   CalculateAngleFromHorizontal centerX  centerY  x2  y2       var angle3   CalculateAngleFromHorizontal centerX  centerY  x3  y3       return  360 0   angle3 - angle2  360        Smaller angle from point2 to point 3 public static double CalculateAngle0To180 double centerX  double centerY  double x2  double y2  double x3  double y3        var angle   CalculateAngle0To360 centerX  centerY  x2  y2  x3  y3       if  angle  gt  180 0                angle   360 - angle            return angle

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Basically what you have is two vectors  one vector from P1 to P2 and another from P1 to P3  So all you need is an formula to calculate the angle between two vectors   Have a look here for a good explanation and the formula

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You mentioned a signed angle  -90   In many applications angles may have signs  positive and negative  see http   en wikipedia org wiki Angle   If the points are  say  P2 1 0   P1 0 0   P3 0 1  then the angle P3-P1-P2 is conventionally positive  PI 2  whereas the angle P2-P1-P3 is negative  Using the lengths of the sides will not distinguish between   and - so if this matters you will need to use vectors or a function such as Math atan2 a  b    Angles can also extend beyond 2 PI and while this is not relevant to the current question it was sufficiently important that I wrote my own Angle class  also to make sure that degrees and radians did not get mixed up   The questions as to whether angle1 is less than angle2 depends critically on how angles are defined  It may also be important to decide whether a line  -1 0  0 0  1 0  is represented as Math PI or -Math PI

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x000D   x000D  function p x  y   return  x y   x000D   x000D  function normaliseToInteriorAngle angle    x000D   if  angle  lt  0    x000D    angle     2 Math PI  x000D     x000D   if  angle  gt  Math PI    x000D    angle   2 Math PI - angle x000D     x000D   return angle x000D    x000D   x000D  function angle p1  center  p2    x000D   const transformedP1   p p1 x - center x  p1 y - center y  x000D   const transformedP2   p p2 x - center x  p2 y - center y  x000D   x000D   const angleToP1   Math atan2 transformedP1 y  transformedP1 x  x000D   const angleToP2   Math atan2 transformedP2 y  transformedP2 x  x000D   x000D   return normaliseToInteriorAngle angleToP2 - angleToP1  x000D    x000D   x000D  function toDegrees radians    x000D   return 360   radians    2   Math PI  x000D    x000D   x000D  console log toDegrees angle p -10  0   p 0  0   p 0  -10     x000D   x000D   x000D

[algorithm] How to calculate an angle from three points?

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