[tex]\begin{gathered} \sin \text{ a =}\frac{5}{13} \\ \cos b=\frac{3}{5} \\ \text{For find the value of cos a :} \\ \text{base}=\sqrt[]{13^2-5^2} \\ b=\sqrt[]{169-25} \\ b=\sqrt[]{144} \\ b=12 \\ \cos \text{ a =}\frac{12}{13} \\ F\in d\text{ the value of sin b:} \\ \text{perpendicular =}\sqrt[]{5^2-3^2} \\ p=\sqrt[]{25-9} \\ p=4 \\ \sin \text{ b =}\frac{4}{5} \\ \cos (a+b)\text{ = }cos\text{ a cos b-sin a sin b} \\ \cos (a+b)=\text{ }\frac{12}{13}\times\frac{3}{5}-\frac{5}{13}\times\frac{4}{5} \\ \cos (a+b)=\frac{36}{65}-\frac{20}{65} \\ \cos (a+b)=\frac{16}{65} \end{gathered}[/tex]
