Answer
(a) $\frac{33}{65}$
(b) $-\frac{56}{65}$
(c) $\frac{63}{16}$
(d) quadrant II.
Work Step by Step
1. Given $sinA=\frac{5}{13}$ with $A$ in quadrant I, form a right triangle of sides $5,12,13$ with angle $A$ facing side $5$, we have $cosA=\frac{12}{13}$ and $tanA=\frac{5}{12}$
2. Given $cosB=-\frac{3}{5}$ with $B$ in quadrantI I, form a right triangle of sides $4,3,5$ with angle $B$ facing side $4$, we have $sinB=\frac{4}{5}$ and $tanB=-\frac{4}{3}$
(a) Use the Addition Formula, we have $sin(A+B)=sinAcosB+cosAsinB=(\frac{5}{13})(-\frac{3}{5})+(\frac{12}{13})(\frac{4}{5})=\frac{33}{65}$
(b) Use the Addition Formula, we have $cos(A+B)=cosAcosB-sinAsinB=(\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})=-\frac{56}{65}$
(c) We have $tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}=\frac{(\frac{5}{12})-(-\frac{4}{3})}{1+(\frac{5}{12})(-\frac{4}{3})}=\frac{63}{16}$
(d) As $sin(A+B)\gt0$ and $cos(A+B)\lt0$, angle $A+B$ is in quadrant II.
