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

Obliczamy cos𝛼 z jedynki trygonometrycznej:

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

${\left(\frac{\sqrt{7}}{4}\right)}^2+{\cos }^2\alpha =1$  

$\frac{7}{16}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{9}{16}$  

$\cos \alpha =\frac{3}{4}>0\vee \cos \alpha =-\frac{3}{4}<0-\text{odrzucamy}$  

Odp.  $\cos \alpha =\frac{3}{4}.$  

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b) Rysunek pomocniczy:

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$\sin \alpha =\frac{a}{c}$  

$\frac{\sqrt{7}}{4}=\frac{a}{c}\mid \cdot c$  

$a=\frac{\sqrt{7}}{4}c$  

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Z tw. Pitagorasa:

$a^2+b^2=c^2$  

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

$\frac{b^2}{a^2+c^2}=\frac{c^2-a^2}{a^2+c^2}=\frac{c^2-{\left(\frac{\sqrt{7}}{4}c\right)}^2}{{\left(\frac{\sqrt{7}}{4}c\right)}^2+c^2}=\frac{c^2-\frac{7}{16}{c}^2}{\frac{7}{16}{c}^2+c^2}=\frac{\frac{9}{16}{c}^2}{\frac{23}{16}{c}^2}=\frac{9}{16}\cdot \frac{16}{23}=\frac{9}{23}$