Law of Sines and Cosines


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Posted by Andrew on October 06, 2002 at 21:36:35:

The book had an example a=21 b=16.7 c=10.3

a^2= b^2 + c^2- 2bc cos(A
(21)^2=(16.7)^2 + (10.3)^2 - 2(16.7)(10.3) cos(A
441= 278.89 + 106.09 - 344.02 cos(A
cos(A= 441 - 278.89-106.09/-344.02
Cos A = -0.1628
A=99 (degrees0 22'

So I substituted a b and c for a=5, b=6, c=7
So then I got a small number like .776 or something. The answer is 44(degrees) 25'

What am I doing wrong?



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