How to calculate an angle from three points? [clos

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.

Answer 1

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 > 0)
            angle = Math.Acos(x / magnitude);

        angle = angle * 180 / Math.PI;
        if (y < 0)
            angle = 360 - angle;

        return angle;
    }

Cheers, Paul

Answer 2

Very Simple Geometric Solution with Explanation

_{Few days ago, a fell into the same problem & 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 ϴ, so we need to find out α and β 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 α = (ay - oy) / (ax - ox)   [m = slope = tan ϴ]   ...(eqn-2)

In the same way, for line OB-

tan β = (by - oy) / (bx - ox)   ...(eqn-3)

Now, we need ϴ = β - α. In trigonometry we have a formula-

tan (β-α) = (tan β + tan α) / (1 - tan β * tan α)   ...(eqn-4)

After replacing the value of tan α (from eqn-2) and tan b (from eqn-3) in eqn-4, and applying simplification we get-

tan (β-α) = ( (ax-ox)*(by-oy)+(ay-oy)*(bx-ox) ) / ( (ax-ox)*(bx-ox)-(ay-oy)*(by-oy) )

So,

ϴ = β-α = tan^(-1) ( ((ax-ox)*(by-oy)+(ay-oy)*(bx-ox)) / ((ax-ox)*(bx-ox)-(ay-oy)*(by-oy)) )

That is it!

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Now, take following figure-

This C# or, Java method calculates the angle (ϴ)-

    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<0){
            angleDeg = 180+angleDeg;
        }

        return angleDeg;
    }

Answer 3

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

Answer 4

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 )

Answer 5

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->c (b)   
    var p1c = Math.sqrt(Math.pow(c.x-p1.x,2)+
                        Math.pow(c.y-p1.y,2)); // p1->c (a)
    var p0p1 = Math.sqrt(Math.pow(p1.x-p0.x,2)+
                         Math.pow(p1.y-p0.y,2)); // p0->p1 (c)
    return Math.acos((p1c*p1c+p0c*p0c-p0p1*p0p1)/(2*p1c*p0c));
}

Bonus: Example with HTML5-canvas

Answer 6

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;