# get angle of a line from horizon

I want to know how to get an angle of a line A-B from horizontal axis X. Other questions in SO do that only between two lines. I'm aware I can always draw second line A-C and calculate but I'm wondering if there's a faster method.

EDIT: I'm very sure I'm not doing a premature optimization.

## Answers

You can use atan for that.

angle = atan((By-Ay)/(Bx-Ax))

private double Angulo(int x1, int y1, int x2, int y2) { double degrees; // Avoid divide by zero run values. if (x2 - x1 == 0) { if (y2 > y1) degrees = 90; else degrees = 270; } else { // Calculate angle from offset. double riseoverrun = (double)(y2 - y1) / (double)(x2 - x1); double radians = Math.Atan(riseoverrun); degrees = radians * ((double)180 / Math.PI); // Handle quadrant specific transformations. if ((x2 - x1) < 0 || (y2 - y1) < 0) degrees += 180; if ((x2 - x1) > 0 && (y2 - y1) < 0) degrees -= 180; if (degrees < 0) degrees += 360; } return degrees; }

If

- The angle is small,
- you can live with small inaccuracies, and
- You can use the angle in radians and not degrees,

then there is a fast solution: Under these conditions, you can assume that tan(a) = a = atan(a), and hence just omit the atan() call.

You could also use arccosine, if your line is in the form [r_x,r_y], where r_x is the change in x and r_y is the change in y.

angle = arccos( r_x/( r_x*r_x + r_y*r_y ) )

It's slightly more opaque, but it's basically the dot product law:

angle = arccos (r . v)

Where r and v are both unit vectors (vectors of length 1). In our case, v is the vector [1,0], and r is

[r_x,r_y] / (r_x^2+r_y^2)

in order to make it a unit vector.

The x-axis is actually a line with equation

y = 0

so you could use the solution you have already.

If you need all four quadrants, Atan2 is more suitable than Atan.

public static int GetAngleBetweenPoints(PointF pt1, PointF pt2) { float dx = pt2.X - pt1.X; float dy = pt2.Y - pt1.Y; int deg = Convert.ToInt32(Math.Atan2(dy, dx) * (180 / Math.PI)); if (deg < 0) { deg += 360; } return deg; }