获取点和原点之间的角度

问题描述：

这个问题可能之前已经回答过，如果可以的话，对不起. 我基本上需要获得从原点到点的角度. 因此，假设来源为(0，0)，而我的目标点为(3，0).

This might have been answered before, sorry if it has. I basically need to get the angle from origin to point. So lets say Origin is (0, 0) and my target point is (3, 0).

3点= 90度

6点= 180度

9点= 270度

12点= 0度

不知何故，我要做一些数学魔术，发现角度为90度(顶部为0). 原点可能会有所不同，因此我需要一个带有两个参数的方法Origin和TargetPoint，它返回度数为double的角度.

Somehow, I gotta do some math magic, and find out that the angle is 90 degrees (Top is 0). The origin can vary, so I need a method with two parameters, Origin, and TargetPoint, which returns double Angle in Degrees.

是的，我意识到这看起来很简短且没有建设性，但是我做了问题尽可能简单易懂.所有其他的问题已关闭-.-

Yea, I realize this looks short and nonconstructive, but I made the question as simple and understandable as possible. All the other questions were closed -.-

谢谢

答

两个点A和B之间的向量为B-A =(B.x-A.x，B.y-A.y).可以使用点积或

The vector between two points A and B is B-A = (B.x-A.x, B.y-A.y). The angle between two vectors can be calculated with the dot product or atan2.

var vector2 = Target - Origin;
var vector1 = new Point(0, 1) // 12 o'clock == 0°, assuming that y goes from bottom to top

double angleInRadians = Math.Atan2(vector2.Y, vector2.X) - Math.Atan2(vector1.Y, vector1.X);

另请参见在矢量之间查找符号角

获取点和原点之间的角度

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