Answered By Emily D

I wish I could see the context of the question you're asking, but I'll take my best stab at what I think your question is asking for (is there an image or more given information on your end? That might help you determine the answer).

So the law of cosines says that

 $c^2=a^2+b^2-2ab\cos\left(C\right)$c²=a²+b²−2abcos(C) 

Now the angle location IS important, it must be between sides a and b. It's okay if you don't know what side a is or what side b is, you can rearrange the formula to solve for it. Just be careful when you're labelling and plugging in your numbers

If you need to solve for either a or b, you can use the quadratic formula and treat the equation like a polynomial:

 $0=a^2-a\cdot2b\cos\left(C\right)+b^2-c^2$0=a²−a·2bcos(C)+b²−c² 

 $a=\frac{-2bcos\left(C\right)\pm\sqrt{\left(2bcos\left(C\right)\right)^2-4\left(1\right)\left(b^2-c^2\right)}}{2\left(1\right)}$a=−2bcos(C)±√(2bcos(C))²−4(1)(b²−c²)2(1)  

 $b=\frac{-2acos\left(C\right)\pm\sqrt{\left(2acos\left(C\right)\right)^2-4\left(1\right)\left(a^2-c^2\right)}}{2\left(1\right)}$b=−2acos(C)±√(2acos(C))²−4(1)(a²−c²)2(1)  

 

So in order to use it to solve for anything, you need at least two side lengths and one angle on your triangle.

If the context you're given in your question doesn't give you any angles or only gives you one side length, it will be difficult (or impossible) to use the cosine law.