Z twierdzenia sinusów:

$\frac{\left|BC\right|}{\sin \left(\left|\angle CAB\right|\right)}=\frac{3\sqrt{2}}{\sin {30}^o}=\frac{3\sqrt{2}}{\frac{1}{2}}=2\cdot 3\sqrt{2}=6\sqrt{2}$   

 

$\frac{\left|AC\right|}{\sin \left(\left|\angle ABC\right|\right)}=6\sqrt{2}$  

$\frac{6}{\sin \beta }=6\sqrt{2}$ $\frac{6}{\sin \beta }=6\sqrt{2}$  

$\sin \beta =\frac{6}{6\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

$\beta ={45}^o$  

 

A więc:

$\left|\angle BCA\right|={180}^o-\left({45}^o+{30}^o\right)={180}^o-{75}^o={105}^o$  

 

$\sin {105}^o=\sin \left({180}^o-{75}^o\right)=\sin {75}^o$  

 

Zatem:

$\frac{\left|AB\right|}{\sin {75}^o}=6\sqrt{2}$  

$\left|AB\right|=6\sqrt{2}\cdot \sin {75}^o$  

 

Ze wzoru na sinus sumy kątów:

$\sin {75}^o=\sin \left({30}^o+{45}^o\right)=\sin {30}^o\cdot \cos {45}^o+\cos {30}^o\cdot \sin {45}^o=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}=\frac{\sqrt{2}+\sqrt{6}}{4}$  

Zatem:

$\left|AB\right|=6\sqrt{2}\cdot \frac{\sqrt{2}+\sqrt{6}}{4}=\frac{12+6\cdot 2\cdot \sqrt{3}}{4}=\frac{12+12\sqrt{3}}{4}=\left(3+3\sqrt{3}\right)\left[\text{cm}\right]$