作AD垂直BC于D,故AD垂直平面BB`C`C,连接DC',故∠AC'D为AC`与平面BB`C`C所成的角；设棱长为a,AD⊥DC',有AC'=√2a,C'D=√5a/2,故cos∠AC'D=C'D/AC'=√5a/2/√2a=√10/4