CODEWITHSUNDEEP
pythonalgorithmmath
luct
luct

Reputation: 433

Angle between points?

I have a triangle (A, B, C) and am trying to find the angle between each pair of the three points.

The problem is that the algorithms I can find online are for determining the angle between vectors. Using the vectors I would compute the angle between the vector that goes from (0, 0) to the point I have, and that doesn't give me the angles inside the triangle.

OK, here's some code in Python after the method on the Wikipedia page and after subtracting the values:

import numpy as np
points = np.array([[343.8998, 168.1526], [351.2377, 173.7503], [353.531, 182.72]])

A = points[2] - points[0]
B = points[1] - points[0]
C = points[2] - points[1]

for e1, e2 in ((A, B), (A, C), (B, C)):
    num = np.dot(e1, e2)
    denom = np.linalg.norm(e1) * np.linalg.norm(e2)
    print np.arccos(num/denom) * 180

That gives me 60.2912487814, 60.0951900475 and 120.386438829, so what am I doing wrong?

Upvotes: 40

Views: 7835

Answers (6)

BBSysDyn
BBSysDyn

Reputation: 4601

Only need to calculate two angles, the third can calculated from the first two.

First angle, take two vectors B-A, C-A, for the second C-B, A-B

points = np.array([[343.8998, 168.1526], [351.2377, 173.7503], [353.531, 182.72]])
A,B,C = points[0], points[1], points[2]

e1 = B-A; e2 = C-A
denom = np.linalg.norm(e1) * np.linalg.norm(e2)
d1 = np.rad2deg(np.arccos(np.dot(e1, e2)/denom))

e1 = C-B; e2 = A-B
denom = np.linalg.norm(e1) * np.linalg.norm(e2)
d2 = np.rad2deg(np.arccos(np.dot(e1, e2)/denom))

d3 = 180-d1-d2

print (d1,d2,d3)

The output is

19.191300537488704 141.6798063583008 19.128893104210505

Upvotes: 1

JoshAdel
JoshAdel

Reputation: 68722

I would use the law of cosines, since you can easily calculate the length of each side of the triangle and then solve for each angles individually.

Upvotes: 4

Jason S
Jason S

Reputation: 189856

There are two errors here.

  • You missed a factor of π when translating from radians to degrees (it's × 180 / π)

  • You have to be careful about the signs of vectors, since they are directed line segments.

If I make these modifications I get a result that makes sense:

import numpy as np
points = np.array([[343.8998, 168.1526], [351.2377, 173.7503], [353.531, 182.72]])

A = points[2] - points[0]
B = points[1] - points[0]
C = points[2] - points[1]

angles = []
for e1, e2 in ((A, B), (A, C), (B, -C)):
    num = np.dot(e1, e2)
    denom = np.linalg.norm(e1) * np.linalg.norm(e2)
    angles.append(np.arccos(num/denom) * 180 / np.pi)
print angles
print sum(angles)

which prints out

[19.191300537488704, 19.12889310421054, 141.67980635830079]
180.0

I'd probably make things more symmetrical and use A, B, C vectors that are cyclic and sum to zero:

import numpy as np
points = np.array([[343.8998, 168.1526], [351.2377, 173.7503], [353.531, 182.72]])

A = points[1] - points[0]
B = points[2] - points[1]
C = points[0] - points[2]

angles = []
for e1, e2 in ((A, -B), (B, -C), (C, -A)):
    num = np.dot(e1, e2)
    denom = np.linalg.norm(e1) * np.linalg.norm(e2)
    angles.append(np.arccos(num/denom) * 180 / np.pi)
print angles
print sum(angles)

which prints out

[141.67980635830079, 19.12889310421054, 19.191300537488704]
180.0

The minus signs in the dot product come because we're trying to get the inside angles.

I'm sorry we drove you away in your time of need, by closing the question.

Upvotes: 40

Davido
Davido

Reputation: 2931

Create three vectors, one from v2 to v1 (v2-v1), one from v3 to v1 (v3-v1), and one from v3 to v2 (v3-v2). Once you have these three vectors, you can use the algorithms you already found along with the fact that all the angles will add to 180 degrees.

Upvotes: 0

btilly
btilly

Reputation: 46497

Suppose that you want the angle at A. Then you need to find the angle between the vector from A to B and the vector from A to C. The vector from A to B is just B-A. (Subtract the coordinates.)

Upvotes: 0

biziclop
biziclop

Reputation: 49804

Alternatively, if you only know the length of the sides of the triangle, you can use the Law of cosines.

Upvotes: 0

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