Re: complex numbers and the law of cosine-



On Mon, 13 Oct 2008, JEMebius wrote:

deadpickle 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?


I am curious about the provenance of these figures.
Are they the sides of a very flat-shaped obtuse triangle as measured
rather sloppily in
the wee small hours under bad lighting?
Or did your a, b and c come from a bad homework problem?

Johan E. Mebius

Not a bad question: it turns out that a suitable correction in the three
sides less than 0.26% of their size makes it a very flat-shaped triangle.

Cheers, ZVK(Slavek).
.

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