Answer
$\frac{63}{65}$
Work Step by Step
$\cos A=\pm\sqrt {1-\sin^{2}A}=\pm\sqrt {1-(-\frac{3}{5})^{2}}=\pm\frac{4}{5}$
As $A$ is in the fourth quadrant, $\cos A$ is positive.
Therefore, $\cos A=\frac{4}{5}$
Now, $\cos B=\pm\sqrt {1-\sin^{2}B}=\pm\sqrt {1-(\frac{12}{13})^{2}}=\pm\frac{5}{13}$
As $B$ is in the second quadrant, $\cos B$ is negative.
Therefore, $\cos B=-\frac{5}{13}$
Recall that $\sin(A+B)=\sin A\cos B+\cos A \sin B$
Then $\sin(A+B)=(-\frac{3}{5})(-\frac{5}{13})+(\frac{4}{5})(\frac{12}{13})=\frac{63}{65}$
