vezi figura
b)ducem AE si CE perpendicular pe VB si AG perbendicular CE⇒cf reciprocei teoremei celor 3 perpendiculare AG perpendicular pe planul CVB⇒
d(A,(CVB))=AG
ΔABC⇒ cf pitagora AC²=CB²+BA²⇒AC=16
ΔVOA⇒cf pitagora AV²=VO²+AO²=16²+8²⇒AV=8√5
ΔAFV⇒cf pitagora VF²=AV²-AF²=(8√5)²-(4√2)²⇒VF=12√2
[tex] A_{ABV} [/tex]=VF·AB/2=AE·VB/2⇒AE=24/√5
ΔEOA ⇒cf pitagora EO²=AE²-AO²=(24/√5)²-8²⇒EO=16/√5
[tex] A_{AEC} [/tex]=EO·AC/2=AG·CE/2⇒(16/√5)·16=AG·24/√5⇒AG=32/3
c)unghiul diedru dintre planele VCB si ABV este <AEC
in triunghiul dreptunghic AGE⇒sinAEG=AG/AE=
[tex] \frac{ \frac{32}{3} }{ \frac{24}{ \sqrt{5} } } [/tex]=[tex] \frac{4 \sqrt{5} }{9} [/tex]
