Geometry

Calculate Angle from Three Points

Lets first understand the problem, consider this figure-

We want to calculate ϴ, where we know A, B & O.

Now, if we want to get ϴ, we need to find out α and β first. For any straight line, we know-

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!

Now, take the following figure-

Following C# or, Java method implements above theory-

    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;
    }