How do you evaluate $(a^2+b^2-c^2)/(2ab)$ when a=4, b=3, c=5?

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Answer 1 · 2016-12-05T03:26:09

The answer is 0/24, which is 0.

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Seems awfully similar to the Law of Cosines . . . Oh well, hope this helps.

How do you evaluate (a^2+b^2-c^2)/(2ab)(a^2+b^2-c^2)/(2ab) when a=4, b=3, c=5?