View Full Version : Trig. Law of Cosines
12-12-2006, 08:36 PM
Show that any triangle with standard labeling...
a^2+b^2+c^2/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c
I don't get it. Can someone please help me.
12-13-2006, 05:00 PM
It might help you to write it correctly: it should be
(a^2+b^2+c^2)/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c
which is rather different from what you wrote. (Notice the first parenthesis pair?)
Since you mention the cosine law in your title, I guess you know that
a^2= b^2+ c^2- 2bc cos(alpha)
b^2= a^2+ c^2- 2ac cos(beta)
c^2= a^2+ b^2- 2ab cos(gamma)
Add those 3 equations, simplify, and divide both sides by 2abc.
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