a)

$\beta ={180}^{\circ }-{90}^{\circ }-{60}^{\circ }={30}^{\circ }$  

 

$\frac{a}{c}=\sin \alpha$  

$\frac{a}{5}=\sin {60}^{\circ }$  

$\frac{a}{5}=\frac{\sqrt{3}}{2}$  

$2a=5\sqrt{3}\mid :2$  

$a=\frac{5}{2}\sqrt{3}\text{cm}$        

 

$\frac{b}{c}=\cos \alpha$  

$\frac{b}{5}=\frac{1}{2}\mid \cdot 5$  

$b=\frac{5}{2}\text{cm}$     

 

b)

$\beta ={180}^{\circ }-{90}^{\circ }-{30}^{\circ }={60}^{\circ }$   

$\frac{a}{c}=\sin \alpha$  

$\frac{10}{c}=\sin {30}^{\circ }$  

$\frac{10}{c}=\frac{1}{2}$  

$20=c$  

$c=20\text{cm}$  

 

$\frac{b}{c}=\cos {30}^{\circ }$  

$\frac{b}{20}=\frac{\sqrt{3}}{2}\mid \cdot 20$  

$b=10\sqrt{3}\text{cm}$          

 

 

c)

Skorzystajmy z twierdzenia Pitagorasa:

$a^2+b^2=c^2$  

${\left(5\sqrt{2}\right)}^2+b^2={10}^2$  

$50+b^2=100$  

$b^2=50$  

$b=\sqrt{50}$  

$b=5\sqrt{2}\text{cm}$  

 

$\sin \alpha =\frac{a}{c}=\frac{5\sqrt{2}}{10}=\frac{\sqrt{2}}{2}$  

$\alpha ={45}^{\circ }$  

 

$\beta ={180}^{\circ }-{90}^{\circ }-{45}^{\circ }={45}^{\circ }$