一个三角形,已知三个点的坐标,就任意夹角的角度?答:设A(a,b) B(c,d) C(e,f)向量AB=(c-a,d-b),向量BC=(e-c,f-d)cosABC=向量AB向量BC/|AB|*|BC| =(c-a,d-b)*(e-c,f-d)/|(c-a,d-b)|*|(e-c,f-d)| 依此类推
已知三个点,用matlab如何求两条直线的夹角答:theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta=acosd(dot([2-3,5-2],[7-3,4-2])/(norm([2-3,5-2])*norm([7-3,4-2])))