Răspuns :
[tex]\it \left.\begin{aligned} \it cosB=cos15^o=\dfrac{\sqrt6+\sqrt2}{4}\\ \\ \\ \it cosB=\dfrac{AB}{BC} \Rightarrow cos B=\dfrac{6}{BC}\end{aligned} \right\} \Rightarrow \dfrac{6}{BC}=\dfrac{\sqrt6+\sqrt2}{4} \Rightarrow BC=\dfrac{24}{\sqrt6+\sqrt2} \Rightarrow \\ \\ \\ \Rightarrow BC=\dfrac{24(\sqrt6-\sqrt2)}{4}=6(\sqrt6-\sqrt2)[/tex]