在直三棱柱ABC-A1B1C1中,AB=3,BC=2,CA=根号5

连结A1C∵AB＝3,BC＝2,CA＝√5∴AB²＝BC²＋AC²∴AC⊥BC∵直三棱柱∴AA1⊥平面ABC∵BC在平面ABC内∴AA1⊥BC∵AC∩AA1＝A∴BC⊥平面AA1C1C∵A1C在平面AA1C1C内∴A1C⊥BC又∵AC⊥BC∴∠ACA1为二面角A-BC...