这道题难在哪里?
∵BB1∥AA1,∴A1C与BB1所成角为∠AA1C=45°
∵AA1⊥面ABCD,∴AA1⊥AC,∴AA1=AC=√2=BB1
勾股定理得B1C=√3,A1B1=1,∵A1B1⊥面BB1C1C,∴∠A1CB1是A1C与面BB1C1C所成角
∴tanA1CB=A1B1/B1C=√3/3,∴∠A1CB=30°