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            • 1. 如图,在四棱锥PABCD中,PA⊥底面ABCD,AB⊥AD,AC⊥CD,∠ABC=60°,PA=AB=BC,E是PC的中点.
              (1)证明:AE⊥平面PCD;
              (2)求二面角A-PD-C的正弦值.
            • 2. 如图,在四棱台ABCD-A1B1C1D1中,底面ABCD是菱形,AA1=A1B1=AB=1,∠ABC=60°,AA1⊥平面ABCD.
              (1)若点M是AD的中点,求证:C1M∥平面AA1B1B;
              (2)棱BC上是否存在一点E,使得二面角E-AD1-D的余弦值为?若存在,求线段CE的长;若不存在,请说明理由.
            • 3.

              如图,正三棱柱\(ABC-A_{1}B_{1}C_{1}\)的底面边长是\(2\),侧棱长是\(\sqrt{3}\),\(D\)是\(AC\)的中点.


              \((\)Ⅰ\()\)求证:\(B_{1}C/\!/\)平面\(A_{1}BD\);

              \((\)Ⅱ\()\)求二面角\(A_{1}-BD-A\)的大小;

              \((\)Ⅲ\()\)在线段\(AA_{1}\)上是否存在一点\(E\),使得平面\(B_{1}C_{1}E⊥\)平面\(A_{1}BD\),若存在,求出\(AE\)的长;若不存在,说明理由.

            • 4. 如图,四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)的所有棱长都相等,\(AC∩BD=O\),\(A_{1}C_{1}∩B_{1}D_{1}=O_{1}\),四边形\(ACC_{1}A_{1}\)和四边形\(BDD_{1}B_{1}\)均为矩形.
              \((\)Ⅰ\()\)证明:\(O_{1}O⊥\)底面\(ABCD\);
              \((\)Ⅱ\()\)若\(∠CBA=60^{\circ}\),求二面角\(C_{1}-OB_{1}-D\)的余弦值.
            • 5.

              如图所示,在棱长为\(2\)的正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(E\)、\(F\)分别为\(A_{1}D_{1}\)和\(CC_{1}\)的中点


              \((1)\)求证:\(EF/\!/\)平面\(ACD_{1}\);

              \((2)\)在棱\(BB_{1}\)上是否存在一点\(P\),使得二面角\(P—AC—B\)的大小为\(30^{\circ}\),若存在,求出\(BP\)的长,若不存在,请说明理由.

            • 6. 在棱长为\(a\)的正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(M\)、\(N\)分别为\(A_{1}B_{1}\),\(CC_{1}\)的中点.
              \((1)\)求\(B\)到平面\(AMN\)的距离;
              \((2)\)求二面角\(B-AM-N\)的余弦值.
            • 7.
              如图,四边形\(ABCD\) \(BDEF\) 均为菱形,设\(AC\) \(BD\) 相交于点\(O\) ,若\(\angle DAB=\angle DBF={{60}^{0}}\) ,且\(FA=FC\)

              \((\)Ⅰ\()\)求证:\(FC\parallel \) \(/\!/\)平面\(EAD\) ;                            

              \((\)Ⅱ\()\)求二面角\(A-FC-B\)的余弦值.

            • 8.

              如图所示,在四棱锥\(P-ABCD\)中,\(PD⊥\)平面\(ABCD\),四边形\(ABCD\)是菱形,\(AC=2\),\(BD=2 \sqrt{3} \),且\(AC\),\(BD\)交于点\(O\),\(E\)是\(PB\)上任意一点.


              \((1)\)求证:\(AC⊥DE\);

              \((2)\)已知二面角\(A-PB-D\)的余弦值为\( \dfrac{ \sqrt{15}}{5} \),若\(E\)为\(PB\)的中点,求\(EC\)与平面\(PAB\)所成角的正弦值.

            • 9.

              如图所示,在直四棱柱\(ABCD\)\(-\)\(A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\) \(C\)\({\,\!}_{1}\)\(D\)\({\,\!}_{1}\)中,已知\(AD=1\),且\(DC\)\(=\)\(DD\)\({\,\!}_{1}=2\)\(AD\)\(=2\)\(AB\)\(AD\)\(⊥\)\(DC\)\(AB\)\(/\!/\)\(DC\)

              \((1)\)设\(E\)\(DC\)的中点,求证:\(D\)\({\,\!}_{1}\)\(E\)\(/\!/\)平面\(A\)\({\,\!}_{1}\)\(BD\)

              \((2)\)求二面角\(A\)\({\,\!}_{1}—\)\(BD\)\(—\)\(C\)\({\,\!}_{1}\)的余弦值.

              \((3)\) 求直线与平面\(C\)\({\,\!}_{1}\)\(BD\)所成角的正弦值.

            • 10.
              \((\)本小题满分\(12\)分\()\)如图,在五面体\(ABCDEF\)中,\(FA⊥\)平面\(ABCD\),\(AD/\!/BC/\!/FE\),\(AB⊥AD\),\(M\)为\(EC\)的中点,\(AF=AB=BC=FE= AD\).

              \((1)\)求异面直线\(BF\)与\(DE\)所成的角的大小;

              \((2)\)证明平面\(AMD\)\(⊥\)平面\(CDE\)

              \((3)\)求二面角\(A\)\(­\)\(CD\)\(­\)\(E\)的余弦值.

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