Przypomnijmy: wszystkie wartości funkcji trygonometrycznych kąta ostrego są dodatnie.

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a)

$\sin \alpha =0,1=\frac{1}{10}$  

Korzystając z jedynki trygonometrycznej mamy:

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

${\left(\frac{1}{10}\right)}^2+{\cos }^2\alpha =1$  

$\frac{1}{100}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{99}{100}|\sqrt{},\cos \alpha >0$  

$\cos \alpha =\frac{\sqrt{99}}{10}=\frac{3\sqrt{11}}{10}$  

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{1}{10}}{\frac{3\sqrt{11}}{10}}=\frac{1}{{\cancel{10}}^1}\cdot \frac{{\cancel{10}}^1}{3\sqrt{11}}=\frac{1}{3\sqrt{11}}=\frac{1\cdot \sqrt{11}}{3\sqrt{11}\cdot \sqrt{11}}=\frac{\sqrt{11}}{3\cdot 11}=\frac{\sqrt{11}}{33}$  

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b)

$\cos \alpha =0,9=\frac{9}{10}$  

Korzystając z jedynki trygonometrycznej mamy:

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

${\sin }^2\alpha +{\left(\frac{9}{10}\right)}^2=1$  

${\sin }^2\alpha +\frac{81}{100}=1$  

${\sin }^2\alpha =\frac{19}{100}|\sqrt{},\sin \alpha >0$  

$\sin \alpha =\frac{\sqrt{19}}{10}$  

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{19}}{10}}{\frac{9}{10}}=\frac{\sqrt{19}}{9}$   

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c)

$\cos \alpha =\frac{1}{6}$    

Korzystając z jedynki trygonometrycznej mamy:

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

${\sin }^2\alpha +{\left(\frac{1}{6}\right)}^2=1$  

${\sin }^2\alpha +\frac{1}{36}=1$  

${\sin }^2\alpha =\frac{35}{36}|\sqrt{},\sin \alpha >0$  

$\sin \alpha =\frac{\sqrt{35}}{6}$  

Zatem: 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}=\sqrt{35}$     

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d)

$\sin \alpha =\frac{\sqrt{3}}{3}$   

Korzystając z jedynki trygonometrycznej mamy:

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

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

$\frac{3}{9}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{6}{9}|\sqrt{},\cos \alpha >0$  

$\cos \alpha =\frac{\sqrt{6}}{3}$

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}}=\frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{18}}{6}=\frac{3\sqrt{2}}{6}=\frac{\sqrt{2}}{2}$     

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e)

$\mathrm{tg}\alpha =\frac{8}{15}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{8}{15}$  

$8\cos \alpha =15\sin \alpha$  

$\cos \alpha =\frac{15}{8}\sin \alpha$    

Korzystając z jedynki trygonometrycznej mamy:

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

${\sin }^2\alpha +\frac{225}{64}{\sin }^2\alpha =1$  

$\frac{289}{64}{\sin }^2\alpha =1|\cdot \frac{64}{289}$

${\sin }^2\alpha =\frac{64}{289}|\sqrt{},\sin \alpha >0$  

$\sin \alpha =\frac{8}{17}$  

Zatem:

$\cos \alpha =\frac{15}{8}\sin \alpha =\frac{15}{8}\cdot \frac{8}{17}=\frac{15}{17}$   

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f) 

$\mathrm{tg}\alpha =\sqrt{5}$  

$\frac{\sin \alpha }{\cos \alpha }=\sqrt{5}$  

$\sin \alpha =\sqrt{5}\cos \alpha$  

Korzystając z jedynki trygonometrycznej mamy:

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

$5{\cos }^2\alpha +{\cos }^2\alpha =1$   

$6{\cos }^2\alpha =1|:6$

${\cos }^2\alpha =\frac{1}{6}|\sqrt{},\cos \alpha >0$  

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

 Zatem:

$\sin \alpha =\sqrt{5}\cos \alpha =\sqrt{5}\cdot \frac{\sqrt{6}}{6}=\frac{\sqrt{30}}{6}$

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g)

$\mathrm{tg}\alpha =\frac{1}{4}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{1}{4}$  

$4\sin \alpha =\cos \alpha$  

Korzystając z jedynki trygonometrycznej:

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

${\sin }^2\alpha +{\left(4\sin \alpha \right)}^2=1$   

${\sin }^2\alpha +16{\sin }^2\alpha =1$  

$17{\sin }^2\alpha =1|:17$  

${\sin }^2\alpha =\frac{1}{17}|\sqrt{},\sin \alpha >0$  

$\sin \alpha =\frac{1}{\sqrt{17}}=\frac{\sqrt{17}}{17}$  

Zatem:

$\cos \alpha =4\sin \alpha =\frac{4\sqrt{17}}{17}$  

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h)

$\mathrm{tg}\alpha =2\sqrt{3}$  

$\frac{\sin \alpha }{\cos \alpha }=2\sqrt{3}$  

$\sin \alpha =2\sqrt{3}\cos \alpha$  

Korzystając z jedynki trygonometrycznej:

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

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

$12{\cos }^2\alpha +{\cos }^2\alpha =1$  

$13{\cos }^2\alpha =1|:13$  

${\cos }^2\alpha =\frac{1}{13}|\sqrt{},\cos \alpha >0$

$\cos \alpha =\frac{1}{\sqrt{13}}=\frac{\sqrt{13}}{13}$  

Zatem:

$\sin \alpha =2\sqrt{3}\cos \alpha =2\sqrt{3}\cdot \frac{\sqrt{13}}{13}=\frac{2\sqrt{39}}{13}$