题目内容
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分析:连接AC,BD,交于点O,可知∠DA1O为DA1与平面C1CA1所成角,从而可得DA1与平面C1CA1所成角的正弦值.
解答:
解:连接AC,BD,交于点O,则BD⊥AC
∵AA1⊥BD,AC∩AA1=A
∴BD⊥平面C1CA1,
连接A1O,则∠DA1O为DA1与平面C1CA1所成角,设为α
∴sinα=
=
故选A.
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∵AA1⊥BD,AC∩AA1=A
∴BD⊥平面C1CA1,
连接A1O,则∠DA1O为DA1与平面C1CA1所成角,设为α
∴sinα=
DO |
DA1 |
1 |
2 |
故选A.
点评:本题重点考查线面角,解题的关键是利用线面垂直,确定线面角,属于基础题.
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