如图,已知点P在正方体ABCD－A1B1C1D1的对角线BD1上,∠PDA=60°.

(1)做辅助线DB cosPDA = cosPDB * cosBDA cosPDA = cos60°= 1/2 cosBDA = cos45°= 根号2/2 得 cosPDB = 根号2/2 PDB = 45°,即 PDD1 = 45° 所以PD与CC1所成角的大小也为45°(2)作DP的延长线交B1D1于E,过E作A1D1的垂线EE1,交A1D1于E1设正方体边长为1则 ED1=1,ED=根号2,EE1=根号2/2所以 DE1 = 根号6/2DP与平面AA1D1D所成角的大小等于∠EDE1(EE1垂直于面AA1D1D)而 cosEDE1 = DE1/DE = 根号3/2所以∠EDE1 = 30°所以所求角为30°