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            • 1.

              已知正方形\(ABCD\)的边长是\(4\),对角线\(AC\)与\(BD\)交于\(0\),将正方形\(ABCD\)沿对角线\(BD\)折成\(60^{\circ}\)的二面角,并给出下面结论:\(①AC⊥BD\);\(②AD⊥CO\);\(③\triangle AOC\)为正三角形;\(④\cos ∠ADC=\dfrac{3}{4}.\)则其中的真命题是                   (    )

              A.\(①③④\)
              B.\(①②④\)
              C.\(②③④\)
              D.\(①②③\)
            • 2.
              如图正方体\(ABCD-A′B′C′D′\)中,二面角\(D′-AB-D\)的大小是\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(45^{\circ}\)
              C.\(60^{\circ}\)
              D.\(90^{\circ}\)
            • 3.

              在三棱锥\(p-ABC\)中,点\(P\)在底面的正投影恰好落在等边\(\Delta ABC\)的边\(AB\)上,点\(P\)到底面\(ABC\)的距离等于底面边长。设\(\Delta PAC\)为底面所成的二面角的大小为\(a\),\(\Delta PBC\)与底面所成的二面角的大小为\(\beta \),则\(\tan (a+\beta )\)的最小值为

              A.       \(\dfrac{3}{4}\sqrt{3}\)
              B.\(\dfrac{2}{5}\sqrt{3}\)
              C.\(-\dfrac{8}{13}\sqrt{3}\)
              D.\(-\dfrac{5}{8}\sqrt{3}\)
            • 4.
                 如图将正方形\(ABCD\)沿对角线\(BD\)折成直二面角\(A-BD-C\),有如下四个结论:


              \(①AC⊥ BD\);
              \(②\triangle ACD\)是等边三角形;
              \(③AB\)与\(CD\)所成的角为\(60^{\circ}\);
              \(④AB\)与平面\(BCD\)所成的角为\(60^{\circ}\).
              其中错误的结论是\((\)    \()\)

              A. \(①\)                           
              B. \(②\)                      
              C. \(③\)                           
              D. \(④\)
            • 5.

              如图所示,在四棱锥\(P - ABCD\)中,侧面\(PAB⊥\)底面\(ABCD\),底面\(ABCD\)为矩形,\(PA=PB\),\(O\)为\(AB\)的中点,\(OD⊥PC\),若\(PD\)与平面\(PAB\)所成的角为\(30^{\circ}\),则二面角\(D - PC - B\)的余弦值是 (    )

              A.\(\dfrac{\sqrt{3}}{3}\)
              B.\(-\dfrac{\sqrt{3}}{3}\)
              C.\(\dfrac{1}{3}\)
              D.\(-\dfrac{1}{3}\)
            • 6.
              二面角\(α-l-β\)等于\(120^{\circ}\),\(A\)、\(B\)是棱\(l\)上两点,\(AC\)、\(BD\)分别在半平面\(α\)、\(β\)内,\(AC⊥l\),\(BD⊥l\),且\(AB=AC=BD=1\),则\(CD\)的长等于\((\)  \()\)
              A.\( \sqrt {2}\)
              B.\( \sqrt {3}\)
              C.\(2\)
              D.\( \sqrt {5}\)
            • 7. 已知一个\(60^{\circ}\)的二面角的棱上有两点\(A\),\(B\),\(AC\),\(BD\)分别是在这个二面角的两个面内垂直于\(AB\)的线段,若\(AB=4\),\(AC=6\),\(BD=8\),则\(CD=(\)  \()\)
              A.\(2 \sqrt {41}\)
              B.\(2 \sqrt {3}\)
              C.\(2 \sqrt {17}\)
              D.\(10\)
            • 8.

              设直线\(l\)与球\(O\)有且只有一个公共点\(P\),从直线\(l\)出发的两个半平面\(\alpha ,\beta \)截球\(O\)的两个截面圆的半径分别为\(1\)和\(\sqrt{3}\),二面角\(\alpha -l-\beta \)的平面角为\(\dfrac{5\pi }{6}\),则球\(O\)的表面积

              A.\(4\pi \)
              B.\(16\pi \)
              C.\(28\pi \)
              D.\(112\pi \)
            • 9. 过正方形 \(ABCD\)的顶点 \(A\),引 \(PA\)\(⊥\)平面 \(ABCD\),若 \(PA\)\(=\) \(BA\),则平面 \(ABP\)和平面 \(CDP\)所成二面角的大小是\((\)  \()\)
              A.\(30^{\circ}\)                                                    
              B.\(45^{\circ}\)
              C.\(60^{\circ}\)                                                    
              D.\(90^{\circ}\)
            • 10.
              将正方形\(ABCD\)沿对角线\(BD\)折成直二面角\(A-BD-C\),有如下四个结论:

              \(①AC⊥BD\);
              \(②\triangle ACD\)是等边三角形;
              \(③AB\)与平面\(BCD\)所成的角为\(60^{\circ}\);
              \(④AB\)与\(CD\)所成的角为\(60^{\circ}\).
              其中错误的结论是\((\)  \()\)
              A.\(①\)
              B.\(②\)
              C.\(③\)
              D.\(④\)
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