Răspuns :
Folosim teorema sinusurilor:
[tex]\it \dfrac{AC}{sinB}=2R \Rightarrow \dfrac{AC}{sin45^o} = 2\cdot6\sqrt2 \Rightarrow AC=2\cdot6\sqrt2\cdot sin45^o =2\cdot6\sqrt2\cdot\dfrac{\sqrt2}{2} =12[/tex]
Folosim teorema sinusurilor:
[tex]\it \dfrac{AC}{sinB}=2R \Rightarrow \dfrac{AC}{sin45^o} = 2\cdot6\sqrt2 \Rightarrow AC=2\cdot6\sqrt2\cdot sin45^o =2\cdot6\sqrt2\cdot\dfrac{\sqrt2}{2} =12[/tex]