Rysunek pomocniczy:

Obliczmy długość przeciwprostokątnej:

$c^2={10}^2+{24}^2$  

$c^2=100+576$  

$c^2=676$  

$c=26$  

 

$\sin \alpha =\frac{10}{26}=\frac{5}{13}$  

 

$\cos \alpha =\frac{24}{26}=\frac{12}{13}$  

 

$\sin \beta =\frac{12}{13}$  

 

$\cos \beta =\frac{5}{13}$  

 

$\frac{\sin \alpha +\cos \beta }{5}-\frac{1}{6}\cdot \sqrt{1-{\sin }^2\alpha }+\sqrt{\frac{\sin \alpha \cdot \cos \beta }{\sin \beta \cdot \cos \alpha }}=$  

 

$=\frac{\frac{5}{13}+\frac{5}{13}}{5}-\frac{1}{6}\cdot \sqrt{1-{\left(\frac{5}{13}\right)}^2}+\sqrt{\frac{\frac{5}{13}\cdot \frac{5}{13}}{\frac{12}{13}\cdot \frac{12}{13}}}=$  

 

$=\frac{\frac{10}{13}}{5}-\frac{1}{6}\cdot \sqrt{1-\frac{25}{169}}+{\sqrt{\left(\frac{\frac{5}{13}}{\frac{12}{13}}\right)}}^2)=$  

 

$=\frac{2}{13}-\frac{1}{6}\cdot \sqrt{\frac{144}{169}}+\sqrt{{\left(\frac{5}{12}\right)}^2}=$  

 

$=\frac{2}{13}-\frac{1}{6}\cdot \frac{12}{13}+\frac{5}{12}=$  

 

$=\frac{2}{13}-\frac{2}{13}+\frac{5}{12}=$  

 

$=\frac{5}{12}$