python已知三角形的顶点坐标，求任一顶点角度

from math import *

def Node_Angle(a, b, c):
    """
    求出夹角和边长
    :param a: a为公共点,
    :return: angle为节点的夹角， ab为边长
    """

    Vbx = b[0] - a[0]
    Vby = b[1] - a[1]
    Vcx = c[0] - a[0]
    Vcy = c[1] - a[1]
    angle_line = (Vbx * Vcx + Vby * Vcy) / sqrt((Vbx * Vbx + Vby * Vby) * (Vcx * Vcx + Vcy * Vcy) + 1e-10)
    # 节点夹角
    angle = acos(angle_line) * 180.0 / pi
    # 边长ab
    ab = float(sqrt(pow((a[0] - b[0]), 2) + pow((a[1] - b[1]), 2)))
    return round(angle, 3), ab

	a = [0, 3]
    b = [0, 0]
    c = [4, 0]
    # ===========================
    print(Node_Angle(a, c, b))
    结果如下：
    (53.13, 5.0)
    # ===========================
    print(Node_Angle(c, a, b))
    结果如下：
    (36.87, 5.0)