$\text{Dla}\alpha \in \left(0^{\circ },{90}^{\circ }\right):$   $\sin \left({90}^{\circ }-\alpha \right)=\cos \alpha$   $\cos \left({90}^{\circ }-\alpha \right)=\sin \alpha$   $\text{tg}\left({90}^{\circ }-\alpha \right)=\text{ctg}\alpha$   $\text{ctg}\left({90}^{\circ }-\alpha \right)=\text{tg}\alpha$     $\text{Dla}\alpha \in ⟨0^{\circ },{180}^{\circ }⟩:$   $\sin \left({180}^{\circ }-\alpha \right)=\sin \alpha$   $\cos \left({180}^{\circ }-\alpha \right)=-\cos \alpha$   $\text{tg}\left({180}^{\circ }-\alpha \right)=-\text{tg}\alpha$   $\text{ctg}\left({180}^{\circ }-\alpha \right)=-\text{ctg}\alpha$

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Korzystając z jedynki trygonometrycznej obliczamy sinα:

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

${\sin }^2\alpha +{0,1}^2=1$  

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

${\sin }^2\alpha +\frac{1}{100}=1$  

${\sin }^2\alpha =\frac{99}{100}$  

$\sin \alpha =\frac{3\sqrt{11}}{10}>0\text{lub}\sin \alpha =-\frac{3\sqrt{11}}{10}<0-\text{odrzucamy}$  

Kąt α jest ostry, więc sinα>0. Wobec tego ujemną wartość sinusa odrzucamy i otrzymujemy:

$\sin \alpha =\frac{3\sqrt{11}}{10}$  

Wówczas:

$\text{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{3\sqrt{11}}{10}}{0,1}=\frac{\frac{3\sqrt{11}}{10}}{\frac{1}{10}}=\frac{3\sqrt{11}}{10}\cdot 10=3\sqrt{11}$  

$\text{ctg}\alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{0,1}{\frac{3\sqrt{11}}{10}}=\frac{\frac{1}{10}}{\frac{3\sqrt{11}}{10}}=\frac{1}{10}\cdot \frac{10}{3\sqrt{11}}=\frac{1}{3\sqrt{11}}=\frac{1}{3\sqrt{11}}\cdot \frac{\sqrt{11}}{\sqrt{11}}=\frac{\sqrt{11}}{33}$  

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$\text{a)}\sin \left({180}^{\circ }-\sin \alpha \right)=\sin \alpha =\frac{3\sqrt{11}}{10}$  

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$\text{b)}\text{tg}\left({180}^{\circ }-\alpha \right)=-\text{tg}\alpha =-3\sqrt{11}$  

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$\text{c)}\cos \left({90}^{\circ }-\alpha \right)=\sin \alpha =\frac{3\sqrt{11}}{10}$  

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$\text{d)}\text{tg}\left({90}^{\circ }-\alpha \right)=\text{ctg}\alpha =\frac{\sqrt{11}}{33}$