Is this problem for studying for the test? Or is it a take-home test?
x is between 270,360 so x is negative of the form (360-x).
cos x= cos -x
therefore, we can take cos x as acute.
but sin x is negative
sin360-x = - sin x
=root3 *5/13 + 12/13*1/2
=5root3 + 12
=1/root2*5/13 + 1/root2*12/13
Your addition of angles formula is the right way to go about this. The restriction on x just tells you the sign of sin(x), where you use Pythagorean theorem to find sin(x) without actually needing to solve for x. I would double-check your trig there, especially the sin(60).