Bonjour
a) EB.EC = ll EB ll * ll EC ll * cos(EB ; EC)
BCE est équilatéral donc (EB ; EC) = 60° = [tex]\pi[/tex]/3
EB.EC = 4 * 4 * cos(60°) = 16 * (1/2) = 8
b) CE.CF = ll CE ll * ll CF ll * cos(CE ; CF)
Pour les mêmes raisons que a), (CE ; CF) = 60°
CE.CF = 4 * 2 * cos(60°) = 8 * (1/2) = 4
c) CD.CE = ll CD ll * ll CE ll * cos(CD ; CE)
(CD ; CE) = 30° = [tex]\pi[/tex]/6 car 90° - 60° = 30°
CD.CE = 4*4*cos([tex]\pi[/tex]/6) = 16 * ([tex]\sqrt{3}[/tex]/2) = 8[tex]\sqrt{3}[/tex]
d) BA.BF = ll BA ll * ll BF ll * cos(BA ; BF)
BA.BF = 4*4*cos(90°) = 16 * 0 = 0
