a) Korzystając z tw. Pitagorasa mamy:

${\sqrt{5}}^2+b^2=5^2$  

$5+b^2=25\mid -5$    

$b^2=20$  

$b=\sqrt{20}$  

$b=2\sqrt{5}$  

 

Zatem:

$\sin \beta =\frac{2\sqrt{5}}{5}$  

$\cos \beta =\frac{\sqrt{5}}{5}$  

$\mathrm{tg}\beta =\frac{2\sqrt{5}}{\sqrt{5}}=2$  

---

b) Korzystając z tw. Pitagorasa mamy:

$1^2+b^2=3^2$  

$1+b^2=9\mid -1$  

$b^2=8$  

$b=\sqrt{8}$  

$b=2\sqrt{2}$  

 

Zatem:

$\sin \beta =\frac{2\sqrt{2}}{3}$  

$\cos \beta =\frac{1}{3}$  

$\mathrm{tg}\beta =\frac{2\sqrt{2}}{1}=2\sqrt{2}$  

---

c) Korzystając z tw. Pitagorasa mamy:

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

$4+b^2=8\mid -4$  

$b^2=4$  

$b=2$   

 

Zatem:

$\sin \beta =\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

$\cos \beta =\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

$\mathrm{tg}\beta =\frac{2}{2}=1$