# How do I compute $2ab\cos C$ given $a,b,C$? Isn't an operator missing there?

I have an equation. For example the law of cosines, $c^2 = a^2 + b^2 - 2ab \cos C$

So I calculate it all and I get something like this: 2500 cos 130. I calculate the cos 130, and get -0.643 Now what? I have 2500 and -0.643. Do I multiply them? Or what?

Thanks.

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Usually in math the convention is that when an operator is missing and arguments are just juxtaposed, the implicit operator is multiplication. –  Yuval Filmus Jun 15 '11 at 18:02
So, $2ab\cos C = 2\cdot a \cdot b \cos(C)$. –  lhf Jun 15 '11 at 18:06
How did you get to the cosine rule without encountering this convention earlier? I'm genuinely curious. –  ShreevatsaR Jun 15 '11 at 18:13
@ShreevatsaR, I never took trig and I'm taking physics. :) –  Dan the Man Jun 15 '11 at 18:28

So, $2abcosC=2⋅a⋅bcos(C)$ - Ihf
Teamwork! $– The Chaz 2.0 Jun 15 '11 at 18:22 oh. duh. thanks. – Dan the Man Aug 18 '11 at 14:32 Well as everyone points out there is a$\text{multiplication}\$ operator. In other words $$2ab\cos{C} = 2 \times a \times b \times \cos{C}$$