Trigonometry Question (need answer fast)

I'm studying for a big test. First answer that makes perfect sense will get the 10 points.

Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5. (Both u and v are in Quadrant II.)

cos(u+v)

Thanks!

sin u=5/13 dat means cosu= - [sqrt{1-(cosu)^2}]= -12/13
cosv= -3/5 so sin u=sqrt[1-(cosv)^2} = 4/5

cos(u+v)=cosucosv-sinusinv
=(-12/13)(-3/5) - (5/13)(4/5)
=16/65

cos(u + v) = cos u. cos v - sin u. sin v

cos u = - 12/13
sin v = 4/5

hence put the values in the abv eqn u l get.. 16/65

cos u=-12/13
sin v=4/5
cos(u+v)=19/65