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Why I'm getting a complex degree value while using "acos"?

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Abb
Abb 2023-6-10
评论： Image Analyst 2023-6-11
Ran in:
% To reproduce the snippet, you can use below code
XYZc = [2489011.31135707, 7440368.1011554, 17.6551714555564];
X0 = 2489018.662
X0 = 2.4890e+06
Y0 = 7440333.989
Y0 = 7.4403e+06
Z0= 10.091
Z0 = 10.0910
dir_vec1 = XYZc-[X0, Y0, Z0];
dir_mag1= norm(dir_vec1);
dir_vec2 = [0,XYZc(2),0]-[X0, Y0, Z0];
dir_mag2= norm(dir_vec2);
alpha = acosd(dir_mag2/dir_mag1)
alpha = 0.0000e+00 + 6.7868e+02i
  1 个评论
Matt J
Matt J 2023-6-10
I have edited your post for you to make your code output visible. It is always advisable to do this so we can see what output you are talking about.

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采纳的回答

Matt J
Matt J 2023-6-10
编辑：Matt J 2023-6-10
Ran in:
Because abs(dir_mag2/dir_mag1) is greater than 1.
abs(dir_mag2/dir_mag1)
ans = 6.9710e+04
  13 个评论
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Abb
Abb 2023-6-11
编辑：Abb 2023-6-11
@Matt J I do not see "Accepted Answer" in this comment! can you please enable it?
Image Analyst
Image Analyst 2023-6-11
It's not in the comments, neither his nor yours. It should be above, at the very first answer post of @Matt J

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