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$\text{a)}$  

$\sin \alpha =\frac{1}{2}$  

$\sin \alpha =\sin {30}^{\circ }$  

$\alpha ={30}^{\circ }$  

oraz   

$\mathrm{ctg}\beta =1$  

$\mathrm{ctg}\beta =\mathrm{ctg}{45}^{\circ }$  

$\beta ={45}^{\circ }$  

 

Obliczmy trzeci kąt tego trójkąta.

$\gamma ={180}^{\circ }-\left({30}^{\circ }+{45}^{\circ }\right)={180}^{\circ }-{75}^{\circ }={105}^{\circ }$  

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$\text{b)}$  

$\sin \alpha =\frac{\sqrt{3}}{2}$  

$\sin \alpha =\sin {60}^{\circ }$  

$\alpha ={60}^{\circ }$  

oraz

$\mathrm{tg}\beta =\sqrt{3}$  

$\mathrm{tg}\beta ={60}^{\circ }$  

$\beta ={60}^{\circ }$  

 

Obliczmy trzeci kąt tego trójkąta.

$\gamma ={180}^{\circ }-\left({60}^{\circ }+{60}^{\circ }\right)={180}^{\circ }-{120}^{\circ }={60}^{\circ }$  

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$\text{c)}$  

$\cos \alpha =\frac{\sqrt{2}}{2}$  

$\cos \alpha =\cos {45}^{\circ }$  

$\alpha ={45}^{\circ }$  

oraz

$\mathrm{ctg}\beta =\frac{\sqrt{3}}{3}$  

$\mathrm{ctg}\beta =\mathrm{ctg}{60}^{\circ }$  

$\beta ={60}^{\circ }$  

 

Obliczmy trzeci kąt tego trójkąta.

$\gamma ={180}^{\circ }-\left({45}^{\circ }+{60}^{\circ }\right)={180}^{\circ }-{105}^{\circ }={75}^{\circ }$  

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$\text{d)}$  

$\mathrm{tg}\alpha =1$  

$\mathrm{tg}\alpha =\mathrm{tg}{45}^{\circ }$  

$\alpha ={45}^{\circ }$  

oraz

$\sin \beta =\frac{\sqrt{2}}{2}$  

$\sin \beta =\sin {45}^{\circ }$  

$\beta ={45}^{\circ }$  

 

Obliczmy trzeci kąt tego trójkąta.

$\gamma ={180}^{\circ }-\left({45}^{\circ }+{45}^{\circ }\right)={180}^{\circ }-{90}^{\circ }={90}^{\circ }$