连结BC1,B1C,相交于M,连结A1M
则BC1⊥B1C,
A1B1⊥平面BCC1B1,
BC1∈平面BCC1B1,
A1B1⊥BC1,
A1B1∩B1C=B1,
故BC1⊥平面A1B1CD,
A1M是A1B在平面A1B1CD上射影,
〈BA1M就是A1B与平面A1B1CD所成角,
设棱长=1,BM=√2/2,
A1B=√2,
sin
连结BC1,B1C,相交于M,连结A1M
则BC1⊥B1C,
A1B1⊥平面BCC1B1,
BC1∈平面BCC1B1,
A1B1⊥BC1,
A1B1∩B1C=B1,
故BC1⊥平面A1B1CD,
A1M是A1B在平面A1B1CD上射影,
〈BA1M就是A1B与平面A1B1CD所成角,
设棱长=1,BM=√2/2,
A1B=√2,
sin