Re: complex numbers and the law of cosine
- From: "Zdislav V. Kovarik" <kovarik@xxxxxxxxxxx>
- Date: Tue, 14 Oct 2008 14:40:54 -0400
On Sun, 12 Oct 2008, fishfry wrote:
In article
<ba9c19db-4f15-4375-abd6-6542f3c7c4d4@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
deadpickle <deadpickle@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?
a + b = 6.908, less than c.
Precisely, and if one calculates all the alleged "cosines", one gets
cos(A) = 2133/2114 > 1
cos(B) = 631/627 > 1
cos(C) = -677/663 < -1
A triple whammy!
A joke from the military service in the late Czechoslovakia: imagine a
less educated but politically dedicated officer lecturing to math
students:
"The projection is calculated using a function called cosine, which has
values between -1 and 1. But in emergency situations, such as during the
Great Patriotic War [the Soviet share of WWII], it attained values up to
5.0 "
In complex functions course, it is a standard exercise to find a
complex number whose cosine is 5 (for example).
Cheers, ZVK(Slavek).
.
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