a) 

$\text{Pole}=20$  

$\text{Pole}=\frac{1}{2}\cdot 5\cdot 10\cdot \sin \alpha$  

$\frac{1}{2}\cdot 5\cdot 10\cdot \sin \alpha =20$  

$25\sin \alpha =20\mid :25$  

$\sin \alpha =\frac{20}{25}$  

$\sin \alpha =\frac{4}{5}\text{i}\alpha \in \left(0^{\circ };{180}^{\circ }\right)$  

$\alpha \approx {53}^{\circ }\text{lub}\alpha \approx {180}^{\circ }-{53}^{\circ }={127}^{\circ }$  

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b) Szkic:

$\text{Pole}=14$  

$\text{Pole}=\frac{1}{2}\cdot 6\cdot 7\cdot \sin \alpha$  

$\frac{1}{2}\cdot 6\cdot 7\cdot \sin \alpha =14$  

$21\sin \alpha =14\mid :21$  

$\sin \alpha =\frac{14}{21}$  

$\sin \alpha =\frac{2}{3}\text{i}\alpha \in \left(0^{\circ };{180}^{\circ }\right)$  

$\alpha \approx {42}^{\circ }\text{lub}\alpha \approx {180}^{\circ }-{42}^{\circ }={138}^{\circ }$  

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c) Szkic:

$\text{Pole}=12$  

$\text{Pole}=\frac{1}{2}\cdot 6\cdot 8\cdot \sin \alpha$  

$\frac{1}{2}\cdot 6\cdot 8\cdot \sin \alpha =12$  

$24\sin \alpha =12\mid :24$  

$\sin \alpha =\frac{12}{24}$  

$\sin \alpha =\frac{1}{2}\text{i}\alpha \in \left(0^{\circ };{180}^{\circ }\right)$  

$\alpha ={30}^{\circ }\text{lub}\alpha ={180}^{\circ }-{30}^{\circ }={150}^{\circ }$