# Procrastinater needs help withv Law of Cosines

1. Feb 8, 2006

### Paindealer

Ok I did this thing on law of cosines and here is what i came up with, but i am pretty sure it is wrong so help me change it to solve for cos(C)

c^2=a^2+b^2-2abcos(C)

so...

Cos(C)=____c^2____
. a^2+b^2-2ab

is that right?

if not how do i make it to equal cos(C)?
(need this by like right now cause i am a procrastinator!!! :P)

2. Feb 8, 2006

### mathmike

wouldnt cos(x) = (a^2 + b^2 - c^2) / 2ab

3. Feb 8, 2006

### Paindealer

probably... i will use that for my hw then, thanks

4. Feb 9, 2006

### HallsofIvy

No!! You both are apparently thinking of

c2= (a2+ b2- 2ab) cos(C)
which it is NOT!! (Notice the parentheses on the right side.)
The cosine law says c2= a2+ b2- 2ab cos(C). (Notice lack of parentheses!)

The problem is not the "cosine", it is simple algebra:

Suppose the problem were c= a+ bx. How would you solve for x?

5. Feb 9, 2006

### ksinclair13

I don't see how mathmike's is wrong, but I may just be stupid.

Isn't

$$cos (C) = \frac{a^2 + b^2 - c^2}{2ab}$$

the same as

$$cos (C) = \frac{c^2 - a^2 - b^2}{-2ab}$$

?

You could take a -1 out of the numerator and then cancel out the negatives, right?