sunqpl
幼苗
共回答了13个问题采纳率:100% 举报
(1)见解析(2)60°
(1)取BD的中点O,在线段CD上取点F,使得DF=3CF,连接OP、OF、FQ
∵△ACD中,AQ=3QC且DF=3CF,∴QF∥AD且QF=
![](https://img.yulucn.com/upload/0/b6/0b6dd8133b69e4ad87f574cba413e78e_thumb.jpg)
AD
∵△BDM中,O、P分别为BD、BM的中点
∴OP∥DM,且OP=
![](https://img.yulucn.com/upload/5/c5/5c5c7f1e7e1bdaeb285be5f53efaf252_thumb.jpg)
DM,结合M为AD中点得:OP∥AD且OP=
![](https://img.yulucn.com/upload/0/b6/0b6dd8133b69e4ad87f574cba413e78e_thumb.jpg)
AD
∴OP∥QF且OP=QF,可得四边形OPQF是平行四边形
∴PQ∥OF
∵PQ⊄平面BCD且OF⊂平面BCD,∴PQ∥平面BCD;
(2)过点C作CG⊥BD,垂足为G,过G作GH⊥BM于H,连接CH
∵AD⊥平面BCD,CG⊂平面BCD,∴AD⊥CG
又∵CG⊥BD,AD、BD是平面ABD内的相交直线
∴CG⊥平面ABD,结合BM⊂平面ABD,得CG⊥BM
∵GH⊥BM,CG、GH是平面CGH内的相交直线
∴BM⊥平面CGH,可得BM⊥CH
因此,∠CHG是二面角C﹣BM﹣D的平面角,可得∠CHG=60°
设∠BDC=θ,可得
Rt△BCD中,CD=BDcosθ=2
![](https://img.yulucn.com/upload/d/fc/dfc7e6e6cbe2e99c1d59fdabe4539fca_thumb.jpg)
cosθ,CG=CDsinθ=
![](https://img.yulucn.com/upload/5/fc/5fc9a89ec721c5b13539ae538f695e89_thumb.jpg)
sinθcosθ,BG=BCsinθ=2
![](https://img.yulucn.com/upload/d/fc/dfc7e6e6cbe2e99c1d59fdabe4539fca_thumb.jpg)
sin
2 θ
Rt△BMD中,HG=
![](https://img.yulucn.com/upload/7/83/783ea1416f1234440b959f2f38d81036_thumb.jpg)
=
![](https://img.yulucn.com/upload/6/ee/6eee116d173eb9f5dba3ce0e2391a536_thumb.jpg)
;Rt△CHG中,tan∠CHG=
![](https://img.yulucn.com/upload/7/a3/7a3b29c06d94ab656ee351060f808d42_thumb.jpg)
=
∴tanθ=
![](https://img.yulucn.com/upload/6/9a/69afe114d35df659c24f8fd213ac59b0_thumb.jpg)
,可得θ=60°,即∠BDC=60°
1年前
10