$a\text{)}\mathrm{tg}\alpha =\frac{3}{4}=0,75$  

$\sin \alpha =\frac{3}{5}=0,6$  

$\cos \alpha =\frac{4}{5}=0,8$  

 

b) Z twierdzenia Pitagorasa

$6^2+b^2={\left(6\sqrt{2}\right)}^2$   

$36+b^2=36\cdot 2\mid -36$  

$b^2=36\text{i}b>0$  

więc

$b=6$  

 

$\mathrm{tg}\alpha =\frac{6}{6}=1$  

$\sin \alpha =\frac{6}{6\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

$\cos \alpha =\frac{6}{6\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

 

c) Z twierdzenia Pitagorasa

${\left(\sqrt{2}\right)}^2+a^2=3^2$   

$2+a^2=9$  

$a^2=7\text{i}a>0$  

więc

$a=\sqrt{7}$  

 

$\mathrm{tg}\alpha =\frac{\sqrt{2}}{\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{14}}{7}$  

$\sin \alpha =\frac{\sqrt{2}}{3}$  

$\cos \alpha =\frac{\sqrt{7}}{3}$