If sin A=3/5, <A is in quadrant 1, cos B= -5/13, and <B is in quadrant two find:
sin(A+B)
for angle A, sketch a reference triangle in quad I ...
since $\displaystyle \sin{A} = \frac{3}{5}$ , opposite side = 3 , hypotenuse = 5 ... so the adjacent side = 4 ...therefore, $\displaystyle \cos{A} = \frac{4}{5}$
sketch another reference triangle in quad II and use the same procedure to find $\displaystyle \sin{B}$ ... remember the side that runs along the x-axis will have a negative value.
finally, sub in your values into the sum identity for sine ...
$\displaystyle \sin(A+B) = \sin{A}\cos{B} + \cos{A}\sin{B}$