How do I find Vector C in a non-right triangle using the Law of Cosines?

[stolencookie]

Homework Statement

Vector A: 1.96N at 20° Vector B: 1.71N at 65°
There are different parts I did the graph part by hand. I am having trouble finding it analytically. We have to use cm for our triangle.

Homework Equations

The Attempt at a Solution

Vector A: 1.96 cm at 20°
Vector B: 1.71 cm at 65°
Vector C: ? at 95°
180°-85°=95°
Then I had trouble after that since it isn't a right triangle. Would I use sine laws or cosine laws? I am lost.

 

[Charles Link]

If I understand the problem correctly, side $A$ makes an angle of 20 degrees with side $C$, and the angle opposite side $C$ is angle $c= 65$ degrees. If that is correct, the law of cosines can be used to find side $C$: $\\$ $C^2=A^2+B^2-2AB \cos(c)$. $\\$ It would help to have a diagram, but I think I got it right.

 

[stolencookie]

If I understand the problem correctly, side $A$ makes an angle of 20 degrees with side $C$, and the angle opposite side $C$ is angle $c= 65$ degrees. If that is correct, the law of cosines can be used to find side $C$: $\\$ $C^2=A^2+B^2-2AB \cos(c)$. $\\$ It would help to have a diagram, but I think I got it right.

Thank you, I wish i knew how to put up a triangle...