求射线夹角:a(x0,y0),b(x1,y1),c(x2,y2)求射线ab和ac的夹角(可能为钝角)

Question

求射线夹角:a(x0,y0),b(x1,y1),c(x2,y2)求射线ab和ac的夹角(可能为钝角)

Answer 1

射线ab的斜率k1=(y1-y0)/(x1-x0)
射线ac的斜率k2=(y2-y0)/(x2-x0)
设射线ab和ac的夹角为θ,则
tanθ=(k2-k1)/(1+k2×k1)
=[(y2-y0)/(x2-x0)-(y1-y0)/(x1-x0)]/[1+(y1-y0)/(x1-x0)×(y2-y0)/(x2-x0)]
=[(y2-y0)(x1-x0)-(y1-y0)(x2-x0)]/[(x2-x0)(x1-x0)+(y2-y0)(y1-y0)]
然后根椐tanθ的值求出θ
所得值为正,则θ为锐角
所得值为负,则θ为钝角
若分母为零,则θ=90°