$a\text{)}\left|AC\right|=\sqrt{6},\left|BC\right|=2,\left|\sphericalangle ABC\right|={60}^{\circ }$  

wykonajmy rysunek pomocniczy:   



$\frac{2}{\sin \alpha }=\frac{\sqrt{6}}{\sin {60}^{\circ }}$  

$\frac{2}{\sin \alpha }=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}$  

$\frac{2}{\sin \alpha }=\frac{2\sqrt{6}}{\sqrt{3}}$  

$2\sqrt{6}\sin \alpha =2\sqrt{3}$  

$\sin \alpha =\frac{2\sqrt{3}}{2\sqrt{6}}$  

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

$\sin \alpha =\frac{\sqrt{18}}{6}$  

$\sin \alpha \approx 0,7$  

$\alpha \approx {45}^{\circ }$  


Obliczmy miarę kąta β:

$\beta ={180}^{\circ }-{60}^{\circ }-{45}^{\circ }={120}^{\circ }-{45}^{\circ }={75}^{\circ }$  

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$b\text{)}\left|AC\right|=1,\left|BC\right|=\sqrt{2},\left|\sphericalangle ABC\right|={30}^{\circ }$  

wykonajmy rysunek pomocniczy:



$\frac{1}{\sin {30}^{\circ }}=\frac{\sqrt{2}}{\sin \delta }$  

$\frac{1}{\frac{1}{2}}=\frac{\sqrt{2}}{\sin \delta }$  

$2=\frac{\sqrt{2}}{\sin \delta }$  

$\sin \delta =\frac{\sqrt{2}}{2}$  

$\delta ={45}^{\circ }$  

Obliczmy miarę kąta 𝛾:

  $\gamma ={180}^{\circ }-{30}^{\circ }-{45}^{\circ }={150}^{\circ }-{45}^{\circ }={105}^{\circ }$