∵BB1//AA1,AA1⊥底面ABC,∴BB1⊥底面ABC,∴平面BB1C1C⊥底面ABC,取BC的中点D,连接AD,DE,∵△ABC是正三角形,∴AD⊥BC,∵平面BB1C1C∩底面ABC=BC∴AD⊥平面BB1C1C,∴∠AED就是AE与平面BB1C1C所成的角,设AB=a,则BB1=AA...