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;
}
```