C++ 计算2个点的角度

给定P1和P2，如何获得P1到P2的角度？谢谢

好的，记得高中三角赛。这就是我得到的

两点是A（x1，y1）和B（x2，y2）

我假设你想要两点和原点之间的角度O（0,0）

每个点都是一个三角形，以它的高度，底面和斜边为界，所以你得到两个角度alpha1和alpha2。这样做的目的是找到其中的每一个，并计算出所需的角度β，方法是：β=alpha1-alpha2，其中alpha1是alpha1>alpha2

计算alpha1=inv_tan（y1/x1）和 alpha2=库存（y2/x2）

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然后做beta=alpha1-alpha2

这只是

浮动角度=atan2（p1.y-p2.y，p1.x-p2.x）

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当然，返回类型以弧度为单位，如果需要以度为单位，只需做一个学究，两点之间没有角度。然而，有两个向量可以。看看点积，看看你能从中得到什么。我相信OP是指连接P1和P2的线与x轴的夹角。至少，选择的答案就是这么做的。其他答案是相同的，但更简洁。

//This working code is for windows hdc mouse coords gives the angle back that is used in windows. It assumes point 1 will be your origin point
// Tested and working on VS 2017 using 2 mouse coordinates in hdc.
//
//code to call our function.
float angler = get_angle_2points(Point1X, Point1Y, Point2X, Point2Y);


// Takes 2 Window coords(points), turns them into vectors using the origin and calculates the angle around the xaxis between them.
// This function can be used for any hdc window. Ie 2 mouse points.
float get_angle_2points(int p1x, int p1y, int p2x,int p2y)
{
    //Make point1 the origin, make point2 relative to the origin so we do point1 - point1, and point2-point1,
    //since we dont need point1 for the equation to work, the equation works correctly with the origin 0,0.
    int deltaY = p2y - p1y;
    int deltaX = p2x - p1x; //Vector 2 is now relative to origin, the angle is the same, we have just transformed it to use the origin.

    float angleInDegrees = atan2(deltaY, deltaX) * 180 / 3.141;

    angleInDegrees *= -1; // Y axis is inverted in computer windows, Y goes down, so invert the angle.

    //Angle returned as:
    //                      90
    //            135                45
    //
    //       180          Origin           0
    //
    //           -135                -45
    //
    //                     -90


    // returned angle can now be used in the c++ window function used in text angle alignment. ie plf->lfEscapement = angle*10;
    return angleInDegrees;
}