1) Given the values of the positive acute angles A and B
2) Let's calculate sin(A+B), let's then find the cos(A)
[tex]\begin{gathered} \sin ^2(A)\text{ +}\cos ^2(A)=1 \\ \cos ^2(A)=1-(\frac{3}{5})^2 \\ \cos ^{}(A)=\sqrt[]{1-\frac{9}{25}} \\ \cos (A)=\frac{4}{5} \end{gathered}[/tex]Now let's find the sin (B), using the same Pythagorean Identity
[tex]\begin{gathered} \sin ^2(B)=1-\cos ^2(B) \\ \sin (B)\text{ =}\sqrt[]{1-(\frac{8}{17})^2} \\ \sin (B)\text{ =}\frac{15}{17} \end{gathered}[/tex]3)Finally, let's calculate the sin (A+B)
[tex]\begin{gathered} \sin (A+B)\text{ =}\sin (A)\cos B+\cos A\sin B \\ \sin (A+B)=\frac{3}{5}\times\frac{8}{17}+\frac{4}{5}\times\frac{15}{17}\text{ =}\frac{84}{85} \end{gathered}[/tex]sin(A+B) = 84/85
