【答案】
分析:法一:(Ⅰ)證明面PAD⊥面PCD,只需證明面PCD內(nèi)的直線CD,垂直平面PAD內(nèi)的兩條相交直線AD、PD即可;
(Ⅱ)過點(diǎn)B作BE∥CA,且BE=CA,∠PBE是AC與PB所成的角,解直角三角形PEB求AC與PB所成的角;
(Ⅲ)作AN⊥CM,垂足為N,連接BN,說明∠ANB為所求二面角的平面角,在三角形AMC中,用余弦定理求面AMC與面BMC所成二面角的大小.
法二:以A為坐標(biāo)原點(diǎn)AD長為單位長度,建立空間直角坐標(biāo)系,
(Ⅰ)求出
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,計(jì)算
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,推出AP⊥DC.,然后證明CD垂直平面PAD,即可證明面PAD⊥面PCD;
(Ⅱ)
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,計(jì)算
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.即可求得結(jié)果.
(Ⅲ)在MC上取一點(diǎn)N(x,y,z),則存在使
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,說明∠ANB為所求二面角的平面角.求出
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,計(jì)算
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即可取得結(jié)果.
解答:
法一:(Ⅰ)證明:∵PA⊥面ABCD,CD⊥AD,
∴由三垂線定理得:CD⊥PD.
因而,CD與面PAD內(nèi)兩條相交直線AD,PD都垂直,
∴CD⊥面PAD.
又CD?面PCD,
∴面PAD⊥面PCD.
(Ⅱ)解:過點(diǎn)B作BE∥CA,且BE=CA,
則∠PBE是AC與PB所成的角.
連接AE,可知AC=CB=BE=AE=

,又AB=2,
所以四邊形ACBE為正方形.由PA⊥面ABCD得∠PEB=90°
在Rt△PEB中BE=a
2=3b
2,PB=
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,
∴
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.
∴AC與PB所成的角為

.
(Ⅲ)解:作AN⊥CM,垂足為N,連接BN.
在Rt△PAB中,AM=MB,又AC=CB,
∴△AMC≌△BMC,
∴BN⊥CM,故∠ANB為所求二面角的平面角
∵CB⊥AC,由三垂線定理,得CB⊥PC,
在Rt△PCB中,CM=MB,所以CM=AM.
在等腰三角形AMC中,AN•MC=

,
∴
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.
∴AB=2,
∴
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故所求的二面角為
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.
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法二:因?yàn)镻A⊥PD,PA⊥AB,AD⊥AB,以A為坐標(biāo)原點(diǎn)AD長為單位長度,
如圖建立空間直角坐標(biāo)系,則各點(diǎn)坐標(biāo)為
A(0,0,0)B(0,2,0),C(1,1,0),
D(1,0,0),P(0,0,1),M
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(Ⅰ)證明:因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212549379831654/SYS201310232125493798316021_DA/16.png">,
故
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,所以AP⊥DC.
又由題設(shè)知AD⊥DC,且AP與與AD是平面PAD內(nèi)的兩條相交直線,由此得DC⊥面PAD.
又DC在面PCD上,故面PAD⊥面PCD
(Ⅱ)解:因
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,
故
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=
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,
所以
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由此得AC與PB所成的角為
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.
(Ⅲ)解:在MC上取一點(diǎn)N(x,y,z),
則存在使
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,
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,
∴x=1-λ,y=1,z=
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λ.
要使AN⊥MC,只需
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即
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,
解得
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.可知當(dāng)
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時(shí),N點(diǎn)坐標(biāo)為
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,能使
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.
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,
有
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由
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得AN⊥MC,BN⊥MC.所以∠ANB為所求二面角的平面角.
∵
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,
∴

.
故所求的二面角為arccos
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.
點(diǎn)評:本題考查平面與平面垂直,二面角的求法,異面直線所成的角,考查空間想象能力,邏輯思維能力,轉(zhuǎn)化思想,是中檔題.