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            • 1.

              如图,四棱锥\(P-ABCD\)中,\(\triangle PAD\)为等边三角形,且平面\(PAD⊥\)平面\(ABCD\),\(AD=2BC=2\),\(AB⊥AD\),\(AB⊥BC\).

              \((1)\)证明:\(PC⊥BC\);

              \((2)\)若直线\(PC\)与平面\(ABCD\)所成角为\(60^{0}\),求二面角\(B-PC-D\)的余弦值.

            • 2.

              如图,在直三棱柱\(A_{1}B_{1}C_{1}—ABC\)中,\(AB⊥AC\),\(AB=AC=2\),\(A_{1}A=4\),点\(D\)是\(BC\)的中点.

              \((1)\)求异面直线\(A_{1}B\)与\(C_{1}D\)所成角的余弦值;

              \((2)\)求平面\(ADC_{1}\)与平面\(ABA_{1}\)所成二面角的正弦值.

            • 3. 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.
            • 4.
              如图,在边长为\(4\)的菱形\(ABCD\)中,\(∠BAD=60^{\circ}\),\(DE⊥AB\)于点\(E\),将\(\triangle ADE\)沿\(DE\)折起到\(\triangle A_{1}DE\)的位置,使\(A_{1}E⊥EB\).

              \((1)\)求证:\(A_{1}D⊥DC\);
              \((2)\)求二面角\(E-A_{1}B-C\)的余弦值;
              \((3)\)判断在线段\(EB\)上是否存在一点\(P\),使平面\(A_{1}DP⊥\)平面\(A_{1}BC\)?若存在,求出\( \dfrac {EP}{EB}\)的值,若不存在,说明理由.
            • 5.
              如图,在边长为\(4\)的菱形\(ABCD\)中,\(∠BAD=60^{\circ}\),\(DE⊥AB\)于点\(E\),将\(\triangle ADE\)沿\(DE\)折起到\(\triangle A_{1}DE\)的位置,使\(A_{1}E⊥EB\).
              \((1)\)求证:\(A_{1}D⊥DC\);
              \((2)\)求直线\(ED\)与平面\(A_{1}BC\)所成角的正弦值;
              \((3)\)求二面角\(E-A_{1}B-C\)的余弦值.
            • 6. 如图,在四棱锥\(P-ABCD\)中,底面\(ABCD\)为菱形,\(∠BAD=60^{\circ}\),\(Q\)为\(AD\)的
              中点.
              \((\)Ⅰ\()\)若\(PA=PD\),求证:平面\(PQB⊥\)平面\(PAD\);
              \((\)Ⅱ\()\)若平面\(PAD⊥\)平面\(ABCD\),且\(PA=PD=AD=2\),点\(M\)在线段\(PC\)上,试
              确定点\(M\)的位置,使二面角\(M-BQ-C\)大小为\(60^{\circ}\),并求出\( \dfrac {PM}{PC}\)的值.
            • 7. 如图为一简单组合体,其底面\(ABCD\)为边长\(2\)正方形,\(PD⊥\)平面\(ABCD\),\(EC/\!/PD\),且\(PD=2 \sqrt {2},CE= \sqrt {2}.\) 
              \((1)\)若\(N\)为线段\(PB\)的中点,求证:\(EN⊥\)平面\(PDB\).
              \((2)\)求平面\(PBE\)与平面\(ABCD\)所成的二面角的大小.
            • 8.
              如图所示,已知四棱锥\(P-ABCD\)中,\(PA⊥\)平面\(ABCD\),底面\(ABCD\)是直角梯形,\(AB/\!/CD\),\(AB⊥AD\),\(AB=2AD=2AP=2CD=2\),\(E\)是棱\(PC\)上一点,且\(CE=2PE\).
              \((1)\)求证:\(AE⊥\)平面\(PBC\);
              \((2)\)求二面角\(A-PC-D\)的大小.
            • 9. 如图所示,在长方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(AA_{1}=AD=1\),\(E\)为\(CD\)的中点.
              \((1)\)求证:\(B_{1}E⊥AD_{1}\)
              \((2)\)若二面角\(A-B_{1}E-A_{1}\)的大小为\(30^{\circ}\),求\(AB\)的长.
            • 10.
              如图,在直角梯形\(ABCD\)中,\(AB⊥AD\),\(AB=AD=2\),\(CD=4\),点\(E\)为\(CD\)中点,将三角形\(ABD\)沿\(BD\)翻折.
              \((\)Ⅰ\()\) 证明:在翻折过程中,始终有\(AE⊥BD\);
              \((\)Ⅱ\()\) 当\(AC=2 \sqrt {3}\)时,求二面角\(A-BD-C\)的大小.
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