Re: complex numbers and the law of cosine
- From: Kobu <kobu.selva@xxxxxxxxx>
- Date: Sun, 12 Oct 2008 00:10:58 -0700 (PDT)
On Oct 12, 3:02 am, deadpickle <deadpic...@xxxxxxxxx> wrote:
I am trying to find an angle in an obtuse triangle. The only way I can
figure to do this is to use the law of cosine. Here's the problem:
c^2=a^2+b^2-2ab cos C
B=arccos((c^2-a^2-b^2)/(-2ab))
were:
a = 3.105
b = 3.803
c = 6.944
When I solve for the angle B I get a complex number (), why?
Visualizing the problem show that the angle should be solvable. Is
there a way to get a real number for this?
Triangle can't exist, because it doesn't satisfy triangle inequality:
a + b >= c (needs to be satisfied)
.
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- complex numbers and the law of cosine
- From: deadpickle
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