[tex]\it BB',\ DC' - drepte\ necoplanare\\ \\ AB'||DC' \Rightarrow \widehat{(BB',\ DC')} = \widehat{(BB',\ AB')}=\widehat{(AB'B)}\\ \\ \Delta BAB'-dreptunghic,\ m{(\hat B)}=90^o \Rightarrow tg \widehat{(AB'B)}= \dfrac{AB}{BB'} =\dfrac{\ 6^{(3}}{9}=\dfrac{2}{3}[/tex]
