qiugen
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共回答了26个问题采纳率:100% 举报
(1)证明:在△PAD中,
由题设,PA=2,AD=2,PD=
![](https://img.yulucn.com/upload/2/2d/22d8c4029f9edc30f4f8c219a0527f97_thumb.jpg)
,
可得PA
2 +AD
2 =PD
2 ,
于是AD⊥PA,
在矩形ABCD中,AD⊥AB,
又PA∩AB=A,
∴AD⊥平面PAB。
(2)过点P作PH⊥AB于H,过H作HE⊥BD于E,
连结PE,
∵AD⊥平面PAB,PH
![](https://img.yulucn.com/upload/a/ec/aecc432aa136744e75d328650ae31eec_thumb.jpg)
平面PAB,
∴AD⊥PH,
又AD∩AB=A,
∴PH⊥平面ABCD,
故HE为PE在平面ABCD内的射影,
由三垂线定理,可知BD⊥PE,
从而∠PEH是二面角P-BD-A的平面角,
由题设,可得
PH=PA·sin60°=
![](https://img.yulucn.com/upload/e/ff/effacc324532881511440de5f7e4ea71_thumb.jpg)
,AH=PA·cos60°=1,
BH=AB-AH=2,BD=
![](https://img.yulucn.com/upload/f/57/f577d1cc16b80cdfb06721717e8b0490_thumb.jpg)
,
HE=
![](https://img.yulucn.com/upload/6/4a/64ae3cd5bd7f7c38e5b69c3fd4b77fbc_thumb.jpg)
,
于是在Rt△PHE中,tan∠PEH=
![](https://img.yulucn.com/upload/b/a6/ba6275c15205d72c41daf48fe66a8e73_thumb.jpg)
,
所以二面角P-BD-A的大小为arctan
![](https://img.yulucn.com/upload/d/d6/dd664a919b2845ab8e3b8b5dc442eed2_thumb.jpg)
。
1年前
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