Rysunek pomocniczy:

$2)$  

${180}^{\circ }=\alpha +\beta +{90}^{\circ }$  

$\beta ={90}^{\circ }-\alpha$  

Zauważmy, że:

$\sin \alpha =\frac{a}{c}$  

$\cos \beta =\frac{a}{c}$  

Zatem:         

$\cos \left({90}^{\circ }-\alpha \right)=\cos \beta =\sin \alpha$  

 

$3)$    

$\mathrm{tg}\alpha =\frac{a}{b}$  

$\mathrm{tg}\left({90}^{\circ }-\alpha \right)=\mathrm{tg}\beta =\frac{b}{a}$   

Zauważmy, że:

$\mathrm{tg}\left({90}^{\circ }-\alpha \right)=\frac{b}{a}=\frac{1}{\frac{a}{b}}=\frac{1}{\mathrm{tg}\alpha }$  

 

$4)$  

$\mathrm{tg}\alpha =\frac{a}{b}$  

$\mathrm{ctg}\left({90}^{\circ }-\alpha \right)=\mathrm{ctg}\beta =\frac{a}{b}$   

Zauważmy, że:

$\mathrm{ctg}\left({90}^{\circ }-\alpha \right)=\frac{a}{b}=\mathrm{tg}\alpha$  

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$\text{a)}$  

$\sin \left({90}^{\circ }-\alpha \right)=\frac{3}{10}$  

$\cos \alpha =\frac{3}{10}$  

Z jedynki trygonometrycznej mamy:

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

${\sin }^2\alpha +{\left(\frac{3}{10}\right)}^2=1$  

${\sin }^2\alpha +\frac{9}{100}=1$  

${\sin }^2\alpha =\frac{91}{100}$  

$\sin \alpha =\frac{\sqrt{91}}{10}$  

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{91}}{10}}{\frac{3}{10}}=\frac{\sqrt{91}}{3}$  

$\mathrm{ctg}\alpha =\frac{3}{\sqrt{91}}=\frac{3\sqrt{91}}{91}$  

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$\text{b)}$      

$\cos \left({90}^{\circ }-\alpha \right)=\frac{\sqrt{3}}{2}$  

$\sin \alpha =\frac{\sqrt{3}}{2}$   

 

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

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

$\frac{3}{4}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{1}{4}$  

$\cos \alpha =\frac{1}{2}$  

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$  

$\mathrm{ctg}\alpha =\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$  

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$\text{c)}$  

$\mathrm{ctg}\left({90}^{\circ }-\alpha \right)=\frac{2}{5}$  

$\mathrm{tg}\alpha =\frac{2}{5}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{2}{5}$  

$\sin \alpha =\frac{2}{5}\cos \alpha$  

 

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

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

$\frac{4}{25}{\cos }^2\alpha +{\cos }^2\alpha =1$  

$\frac{29}{25}{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{25}{29}$  

$\cos \alpha =\frac{5}{\sqrt{29}}$  

$\cos \alpha =\frac{5\sqrt{29}}{29}$  

Zatem:

$\sin \alpha =\frac{2}{5}\cdot \frac{5\sqrt{29}}{29}$  

$\sin \alpha =\frac{2\sqrt{29}}{29}$  

 

$\mathrm{tg}\alpha =\frac{2}{5}$  

$\mathrm{ctg}\alpha =\frac{5}{2}$