Przypomnijmy, że:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

$\frac{\sin \alpha }{\cos \alpha }=\mathrm{tg}\alpha$  

 

a)

${\left(\frac{5}{13}\right)}^2+{\cos }^2\alpha =1$  

$\frac{25}{169}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{144}{169}$  

$\cos \alpha =\frac{12}{13}$  

 

$\mathrm{tg}\alpha =\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{13}\cdot \frac{13}{12}=\frac{5}{12}$          

 

b)

${\sin }^2\alpha +{0,6}^2=1$   

${\sin }^2\alpha +0,36=1$  

${\sin }^2\alpha =0,64$  

$\sin \alpha =0,8$  

 

$\mathrm{tg}\alpha =\frac{0,8}{0,6}=\frac{8}{6}=\frac{4}{3}$