# OOps my mistake.....You were correct...

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Posted by Subhotosh Khan on October 08, 2002 at 10:05:06:

In Reply to: Mr G...I think there is a problem with "sign"... posted by Subhotosh Khan on October 08, 2002 at 09:31:56:

: : : 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?

: : I see nothing wrong with what you did. In the previous posts, be careful as the poster did not write the "new" equation using proper notation. That is, do what you did to isolate cos(A), then use the inverse cosine function to determine A.

: : If you want to solve for cos(A) first, that is o.k. too. But it is:
:
: The original equation is:

: a2 = b2 + c2 - 2bc*cos(A)
:

: cos(A) = [a2-b2-c2]/[-2bc]

: : A = cos-1 {[a2-b2-c2]/[-2bc]}

: *********************************************

: : I'm sure this is what she meant.

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