Re: complex numbers and the law of cosine



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?

In a triangle a, b, c you must have a + b > c. Now check your numbers.

Bob Kolker
.



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