How to calculate the coordinates of a arrowhead based on the arrow?

algorithm

Solution

You don't need trig., just some vector arithmetic...

Say the line goes from A to B, with the front vertex of the arrowhead at B. The length of the arrowhead is h = 10(√3) and its half-width is w = 10. We'll denote the unit vector from A to B as U = (B - A)/|B - A| (i.e., the difference divided by the length of the difference), and the unit vector perpendicular to this as V = [-Uy, Ux].

From these quantities, you can calculate the two rear vertices of the arrowhead as B - hU ± wV.

In C++:

struct vec { float x, y; /* … */ };

void arrowhead(vec A, vec B, vec& v1, vec& v2) {
    float h = 10*sqrtf(3), w = 10;
    vec U = (B - A)/(B - A).length();
    vec V = vec(-U.y, U.x);
    v1 = B - h*U + w*V;
    v2 = B - h*U - w*V;
}

If you want to specify different angles, then you will need some trig. to calculate different values of `h` and `w`. Assuming you want an arrowhead of length h and tip-angle θ, then w = h tan(θ/2). In practice, however, it's simplest to specify `h` and `w` directly.

Problem

I have a line that is based on two (x,y) coordinates I know. This line has a starting and an end point. Now I want to add an arrowhead at the end point of the line. I know that the arrow is an equilateral triangle, and therefore each angle has 60 degrees. Additionally, I know the length of one side, which will be 20. I also no one edge of the triangle (that is the end point of the line). How can I calculate the other two points of the triangle? I know I should use some trigonometry but how? P.s. The endpoint of the line should be the arrowhead's tip.

Original source