a) 𝛼 jest kątem ostrym, więc sin𝛼 > 0.

Obliczamy sin𝛼 z jedynki trygonometrycznej:

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

${\sin }^2\alpha +{\left(\frac{\sqrt{5}}{3}\right)}^2=1$  

${\sin }^2\alpha +\frac{5}{9}=1$  

${\sin }^2\alpha =\frac{4}{9}$  

$\sin \alpha =\frac{2}{3}>0\vee \sin \alpha =-\frac{2}{3}<0-\text{odrzucamy}$  

Odp.  $\sin \alpha =\frac{2}{3}.$  

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b) Rysunek pomocniczy:

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$\cos \alpha =\frac{b}{c}$  

$\frac{\sqrt{5}}{3}=\frac{b}{c}\mid \cdot c$  

$b=\frac{\sqrt{5}}{3}c$  

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Z tw. Pitagorasa:

$a^2+b^2=c^2$  

$a^2=c^2-b^2\mid :\left(2b^2+c^2\right)$  

$\frac{a^2}{2b^2+c^2}=\frac{c^2-b^2}{2b^2+c^2}=\frac{c^2-{\left(\frac{\sqrt{5}}{3}c\right)}^2}{2{\left(\frac{\sqrt{5}}{3}c\right)}^2+c^2}=\frac{c^2-\frac{5}{9}{c}^2}{2\cdot \frac{5}{9}{c}^2+c^2}=\frac{\frac{4}{9}{c}^2}{\frac{10}{9}{c}^2+c^2}=\frac{\frac{4}{9}{c}^2}{\frac{19}{9}{c}^2}=\frac{4}{9}\cdot \frac{9}{19}=\frac{4}{19}$