Trig. Law of Cosines

[ashleebrook]

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.

[HallsofIvy]

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.