给定三角形在二维平面上所有三个顶点的坐标,任务是找到所有三个角度。
示例:
输入:A = (0, 0),
B = (0, 1),
C = (1, 0)
输出:90, 45, 45
为了解决这个问题,我们使用下面的余弦定律。
c^2 = a^2 + b^2 - 2(a)(b)(cos beta)
重新安排后
beta = acos( ( a^2 + b^2 - c^2 ) / (2ab) )
在三角学中,余弦定律(也称为余弦公式或余弦规则)将三角形边的长度与其某个角的余弦联系起来。
首先,计算所有边的长度。然后应用上述公式得到所有角度的弧度。然后将角度从弧度转换为度数。
以下是上述步骤的实施:
# Python3 code to find all three angles
# of a triangle given coordinate
# of all three vertices
import math
# returns square of distance b/w two points
def lengthSquare(X, Y):
xDiff = X[0] - Y[0]
yDiff = X[1] - Y[1]
return xDiff * xDiff + yDiff * yDiff
def printAngle(A, B, C):
# Square of lengths be a2, b2, c2
a2 = lengthSquare(B, C)
b2 = lengthSquare(A, C)
c2 = lengthSquare(A, B)
# length of sides be a, b, c
a = math.sqrt(a2);
b = math.sqrt(b2);
c = math.sqrt(c2);
# From Cosine law
alpha = math.acos((b2 + c2 - a2) /
(2 * b * c));
betta = math.acos((a2 + c2 - b2) /
(2 * a * c));
gamma = math.acos((a2 + b2 - c2) /
(2 * a * b));
# Converting to degree
alpha = alpha * 180 / math.pi;
betta = betta * 180 / math.pi;
gamma = gamma * 180 / math.pi;
# printing all the angles
print("alpha : %f" %(alpha))
print("betta : %f" %(betta))
print("gamma : %f" %(gamma))
# Driver code
A = (0, 0)
B = (0, 1)
C = (1, 0)
printAngle(A, B, C);
# This code is contributed
# by ApurvaRaj
输出:
alpha : 90
beta : 45
gamma : 45
时间复杂度:由于使用内置 sqrt 函数,因此为 O(log(n))
辅助空间: O(1)
参考:
https://en.wikipedia.org/wiki/Law_of_cosines
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