Cosine law question. how is it 131 degrees?

2008-02-05 12:07 am
q^2 = r^2 + p^2 - 2rp cos Q

q^2 - r^2 - p^2 = -2rp cos Q

( q^2 - r^2 - p^2 over -2rp ) = cos Q

Q= cos -1 ( q^2 - r^2 - p^2 over -2rp )

Q=cos -1 ( (12.4)^2 - (6.3)^2 - (7.3)^2 over -2(6.3)(7.3) )

Q= 101 degrees


I'm not getting anything close to 101. Can soembody maybe tell me what to punch in because maybe i'm putting it in wrong.
更新1:

sorry. Q= 131 not 101.

回答 (1)

2008-02-05 2:08 am
✔ 最佳答案
Try step by step on the calculator, might be a missed order of operation?
12.4^2 - 6.3^2 - 7.3^2 = 60.78
-2*6.3*7.3 = -91.98
60.78 / (-91.98) ~= -0.6608
So, Q = cos^(-1) ( -0.6608 ) = 131 degrees.


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