a)

$c^2=a^2+b^2-2ab\cos \gamma \text{|}-a^2-b^2$

$c^2-a^2-b^2=-2ab\cos \gamma \text{|}:(-2ab)$

$\frac{c^2-a^2-b^2}{-2ab}=\cos \gamma$  

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

$\cos \gamma =\frac{2+\sqrt{3}-1-1}{-2}$  

$\cos \gamma =\frac{\sqrt{3}}{-2}$  

$-\cos \gamma =\frac{\sqrt{3}}{2}$  

$\cos \left({180}^{\circ }-\gamma \right)=\frac{\sqrt{3}}{2}$  

Wiemy, że:

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$  

Zatem:

${180}^{\circ }-\gamma ={30}^{\circ }$  

$\gamma ={150}^o$  

Możemy zauważyć, że  $a=b=1$  więc trójkąt ten jest równoramienny.

$\alpha +\beta +{150}^o={180}^o$  

$\alpha +\beta ={30}^o$  

Wobec tego  $\alpha =\beta ={15}^o$  

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b)

$c^2=a^2+b^2-2ab\cos \gamma$  

$c^2-a^2-b^2=-2ab\cos \gamma$

$\frac{c^2-a^2-b^2}{-2ab}=\cos \gamma$  

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

$\cos \gamma =\frac{1+2\sqrt{3}+3-2-4}{-4\sqrt{2}}$  

$\cos \gamma =\frac{2\sqrt{3}-2}{-4\sqrt{2}}$  

$\cos \gamma =\frac{\sqrt{3}-1}{-2\sqrt{2}}$  

$\cos \gamma \approx -0,2588$  

$\cos \gamma \approx -\sin {15}^o$  

$-\sin {15}^o=\cos \left({90}^{\oplus }{15}^o\right)=\cos {105}^o$  

$\gamma \approx {105}^o$  

$a^2=b^2+c^2-2bc\cos \alpha$  

$\frac{a^2-b^2-c^2}{-2bc}=\cos \alpha$  

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

$\cos \alpha =\frac{2-4-\left(1+2\sqrt{3}+3\right)}{-4\left(1+\sqrt{3}\right)}$  

$\cos \alpha =\frac{-2-1-2\sqrt{3}-3}{-4\left(1+\sqrt{3}\right)}$  

$\cos \alpha =\frac{-6-2\sqrt{3}}{-4\left(1+\sqrt{3}\right)}$  

$\cos \alpha =\frac{3+\sqrt{3}}{2+2\sqrt{3}}$  

$\cos \alpha \approx 0,8660$  

$\alpha \approx {30}^o$  

$\beta \approx {180}^o-\left({105}^o+{30}^o\right)$  

$\beta \approx {45}^o$