Our approach is to use SOHCAHTOA to derive values for sine and cosines of both A and B.
[tex]\begin{gathered} \sin A=\frac{3}{5},\text{ cosA=}\frac{\sqrt[]{5^2-3^2}}{5}=\frac{4}{5} \\ \cos B=\frac{20}{29},\text{ sinB=}\frac{\sqrt[]{29^2-20^2}}{29}=\frac{21}{29} \end{gathered}[/tex][tex]\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1-\text{tanAtanB}}\text{ WHERE} \\ \tan A=\frac{\sin A}{\cos A},\tan B=\frac{\sin B}{\cos B} \end{gathered}[/tex][tex]\begin{gathered} \tan (A+B)=\frac{\frac{\frac{3}{5}}{\frac{4}{5}}+\frac{\frac{21}{29}}{\frac{20}{29}}}{1-\frac{\frac{3}{5}}{\frac{4}{5}}\times\frac{\frac{21}{29}}{\frac{20}{29}}}=\frac{\frac{3}{4}+\frac{21}{20}}{1-\frac{3}{4}\times\frac{21}{20}}=\frac{\frac{9}{5}}{1-\frac{63}{80}}=\frac{\frac{9}{5}}{\frac{17}{80}} \\ \tan (A+B)=8.47 \end{gathered}[/tex]tan (A+B) = 8.47
