设AB=a延长AE,A1C1,交点F;作FG⊥A1B1延长线于G作EH//A1F所求平面夹角θ即平面EB1F和平面B1C1F的夹角EB1=√5a/2,B1C1=a,EC1=a/2,C1F=√2aB1F=√5a,AB1=√2a,AF=3a,EF=3a/2在△EB1F中,作EM⊥B1F于Mcos∠EFB1=2√5/5EM=...