di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$<img_2998>
$(2)求直线AB与平面A_{1}DC_{1}所成角的余弦值；$

Solution:
$取AD的中点O,连接A_{1}O,因为AA_{1}=A_{1}D，O为AD的中点，所以A_{1}O\perp AD，又平面AA_{1}D_{1}D\perp 平面ABCD,平面AA_{1}D_{1}D\cap 平面ABCD=AD，A_{1}O\subset 平面AA_{1}D_{1}D, 所以A_{1}O\perp 平面ABCD,以点O为坐标原点，\overrightarrow{OD},\overrightarrow{OA_1}的方向分别为y轴，z轴的正方向，建立如图所示的空间直角坐标系.(5分)$<img_3110>$则A(0,-1,0)，B(2,-1,0)， A_{1}(0,0,\sqrt{3}),C_{1}(2,2,\sqrt{3})，D(0,1,0),C(2，1，0)，所以\overrightarrow{AB}=(2，0,0)，\overrightarrow{A_1C_1}=(2,2,0)，\overrightarrow{A_1D}=(0,1,-\sqrt{3})，设平面A_{1}C_{1}D的法向量为n_{1}=(x_{1},y_{1},z_{1}),由\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{A_1C_1}=2x_1+2y_1=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{A_1D}=y_1-\sqrt{3}z_1=0,\end{matrix}\right.令z_{1}=1,则n_{1}=(-\sqrt{3}，\sqrt{3}，1)，(7分)设直线AB与平面A_{1}DC_{1}所成角为\theta _{1}，则\\sin   \theta _{1}=|\\cos  \langle \overrightarrow{AB},n_{1}\rangle |=\frac{|\overrightarrow{AB}\cdot \boldsymbol{n}_1|}{|\overrightarrow{AB}||\boldsymbol{n}_1|}=\frac{\sqrt{21}}{7}，所以\\cos   \theta _{1}=\sqrt{1-\\\sin   ^2\theta _1}=\frac{2\sqrt{7}}{7},则直线AB与平面A_{1}DC_{1}所成角的余弦值为\frac{2\sqrt{7}}{7}.(10分)$

Final answer: $\frac{2\sqrt{7}}{7}$
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$![Image](https://cdntk.53zaixian.com/tq/1701760609782321153/img/compression/M_bafb81f07cd6d372ad69314fd0614640.jpg)
$(3)求平面A_{1}BC与平面A_{1}DC_{1}的夹角的正弦值.$

Solution:
$设平面A_{1}BC的法向量为n_{2}=(x_{2},y_{2},z_{2})，由(2)知平面A_{1}DC_{1}的一个法向量为n_{1}=(-\sqrt{3}，\sqrt{3}，1)，又\overrightarrow{A_1B}=(2,-1,-\sqrt{3}),\overrightarrow{BC}=(0,2,0)，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{A_1B}=2x_2-y_2-\sqrt{3}z_2=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BC}=2y_2=0,\end{matrix}\right.令x_{2}=\sqrt{3},则n_{2}=(\sqrt{3}，0，2)，(12分)设平面A_{1}BC与平面A_{1}DC_{1}的夹角为\theta _{2}，则\\cos   \theta _{2}=|\\cos  \langle n_{1}，n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{|-3+2|}{\sqrt{7}\times \sqrt{7}}=\frac{1}{7}，所以\\sin   \theta _{2}=\sqrt{1-\\\cos   ^2\theta _2}=\frac{4\sqrt{3}}{7},所以平面A_{1}BC与平面A_{1}DC_{1}的夹角的正弦值为\frac{4\sqrt{3}}{7}.(15分)$

Final answer: $\frac{4\sqrt{3}}{7}$
Subfield: Solid Geometry


Question: $如图,在四棱柱ABCD-A_{1}B_{1}C_{1}D_{1}中,底面ABCD是正方形,平面A_{1}ADD_{1}\perp 平面ABCD,AD=AA_{1}=A_{1}D=2.$<img_2998>
$(3)求平面A_{1}BC与平面A_{1}DC_{1}的夹角的正弦值.$

Solution:
$设平面A_{1}BC的法向量为n_{2}=(x_{2},y_{2},z_{2})，由(2)知平面A_{1}DC_{1}的一个法向量为n_{1}=(-\sqrt{3}，\sqrt{3}，1)，又\overrightarrow{A_1B}=(2,-1,-\sqrt{3}),\overrightarrow{BC}=(0,2,0)，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{A_1B}=2x_2-y_2-\sqrt{3}z_2=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{BC}=2y_2=0,\end{matrix}\right.令x_{2}=\sqrt{3},则n_{2}=(\sqrt{3}，0，2)，(12分)设平面A_{1}BC与平面A_{1}DC_{1}的夹角为\theta _{2}，则\\cos   \theta _{2}=|\\cos  \langle n_{1}，n_{2}\rangle |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{|-3+2|}{\sqrt{7}\times \sqrt{7}}=\frac{1}{7}，所以\\sin   \theta _{2}=\sqrt{1-\\\cos   ^2\theta _2}=\frac{4\sqrt{3}}{7},所以平面A_{1}BC与平面A_{1}DC_{1}的夹角的正弦值为\frac{4\sqrt{3}}{7}.(15分)$

Final answer: $\frac{4\sqrt{3}}{7}$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$<img_1488>
$(1)求异面直线EF与A_{1}D所成角的余弦值;$

Solution:
$如图所示，以A为坐标原点，建立空间直角坐标系，设AB=1,则D(0,2,0),F(1,2,1),A_{1}(0,0,4),E\left(1,\frac{3}{2},0\right).$<img_2239>$易得\overrightarrow{EF}=\left(0,\frac{1}{2},1\right)，\overrightarrow{A_1D}=(0,2,-4)，
(2分)
于是\\cos  \langle \overrightarrow{EF}，\overrightarrow{A_1D}\rangle =\frac{\overrightarrow{EF}\cdot \overrightarrow{A_1D}}{|\overrightarrow{EF}||\overrightarrow{A_1D}|}=-\frac{3}{5}.
(4分)
所以异面直线EF与A_{1}D所成角的余弦值为\frac{3}{5}.
(5分)$

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$![Image](https://cdntk.53zaixian.com/tq/1701760610528907267/img/compression/M_aae68a6d0648e88336ff3e745877e257.jpg)
$(1)求异面直线EF与A_{1}D所成角的余弦值;$

Solution:
$如图所示，以A为坐标原点，建立空间直角坐标系，设AB=1,则D(0,2,0),F(1,2,1),A_{1}(0,0,4),E\left(1,\frac{3}{2},0\right).$![Image](https://cdntk.53zaixian.com/tq/1701760610528907267/img/compression/M_8384bc5a7147de10543d3e036799950f.jpg)$易得\overrightarrow{EF}=\left(0,\frac{1}{2},1\right)，\overrightarrow{A_1D}=(0,2,-4)，
(2分)
于是\\cos  \langle \overrightarrow{EF}，\overrightarrow{A_1D}\rangle =\frac{\overrightarrow{EF}\cdot \overrightarrow{A_1D}}{|\overrightarrow{EF}||\overrightarrow{A_1D}|}=-\frac{3}{5}.
(4分)
所以异面直线EF与A_{1}D所成角的余弦值为\frac{3}{5}.
(5分)$

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$<img_1488>
$(2)证明：AF\perp 平面A_{1}ED;$

Solution:
$证明:易知\overrightarrow{AF}=(1,2,1),\overrightarrow{EA_1}=\left(-1,-\frac{3}{2},4\right)，\overrightarrow{ED}=\left(-1,\frac{1}{2},0\right)，因为\overrightarrow{AF}\cdot \overrightarrow{EA_1}=0，\overrightarrow{AF}\cdot \overrightarrow{ED}=0，
(6分)
所以AF\perp EA_{1},AF\perp ED,
(8分)
又EA_{1}\cap ED=E，所以AF\perp 平面A_{1}ED.
(9分)$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$![Image](https://cdntk.53zaixian.com/tq/1701760610528907267/img/compression/M_aae68a6d0648e88336ff3e745877e257.jpg)
$(2)证明：AF\perp 平面A_{1}ED;$

Solution:
$证明:易知\overrightarrow{AF}=(1,2,1),\overrightarrow{EA_1}=\left(-1,-\frac{3}{2},4\right)，\overrightarrow{ED}=\left(-1,\frac{1}{2},0\right)，因为\overrightarrow{AF}\cdot \overrightarrow{EA_1}=0，\overrightarrow{AF}\cdot \overrightarrow{ED}=0，
(6分)
所以AF\perp EA_{1},AF\perp ED,
(8分)
又EA_{1}\cap ED=E，所以AF\perp 平面A_{1}ED.
(9分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$<img_1488>
$(3)求二面角A_{1}-ED-F的正弦值.$

Solution:
$设平面EFD的法向量为u=(x,y,z),则\left\{\begin{matrix}\boldsymbol{u}\cdot \overrightarrow{EF}=0,\\ \boldsymbol{u}\cdot \overrightarrow{ED}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{1}{2}y+z=0,\\ -x+\frac{1}{2}y=0,\end{matrix}\right.不妨令x=1,可得u=(1,2,-1).
(11分)
由(2)可知\overrightarrow{AF}为平面A_{1}ED的一个法向量，于是\\cos  \langle u，\overrightarrow{AF}\rangle =\frac{\boldsymbol{u}\cdot \overrightarrow{AF}}{|\boldsymbol{u}|\cdot |\overrightarrow{AF}|}=\frac{2}{3}，
(13分)
从而\\sin  \langle u，\overrightarrow{AF}\rangle =\frac{\sqrt{5}}{3}，所以二面角A_{1}-ED-F的正弦值为\frac{\sqrt{5}}{3}.
(15分)$

Final answer: $\frac{\sqrt{5}}{3}$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E、F分别是棱BC,CC_{1}上的点，CF=AB=2CE,AB:AD:AA_{1}=1:2:4.$![Image](https://cdntk.53zaixian.com/tq/1701760610528907267/img/compression/M_aae68a6d0648e88336ff3e745877e257.jpg)
$(3)求二面角A_{1}-ED-F的正弦值.$

Solution:
$设平面EFD的法向量为u=(x,y,z),则\left\{\begin{matrix}\boldsymbol{u}\cdot \overrightarrow{EF}=0,\\ \boldsymbol{u}\cdot \overrightarrow{ED}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{1}{2}y+z=0,\\ -x+\frac{1}{2}y=0,\end{matrix}\right.不妨令x=1,可得u=(1,2,-1).
(11分)
由(2)可知\overrightarrow{AF}为平面A_{1}ED的一个法向量，于是\\cos  \langle u，\overrightarrow{AF}\rangle =\frac{\boldsymbol{u}\cdot \overrightarrow{AF}}{|\boldsymbol{u}|\cdot |\overrightarrow{AF}|}=\frac{2}{3}，
(13分)
从而\\sin  \langle u，\overrightarrow{AF}\rangle =\frac{\sqrt{5}}{3}，所以二面角A_{1}-ED-F的正弦值为\frac{\sqrt{5}}{3}.
(15分)$

Final answer: $\frac{\sqrt{5}}{3}$
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$<img_3120>
$(1)求证:DE\perp 平面SCD;$

Solution:
$证明：以C为原点建立空间直角坐标系，如图所示，$<img_2799>$则C(0，0，0)，E(0，2，0),S(0，0，3),A\left(\frac{3}{2},0,0\right)，D(1，1，0),B(0，3，0)，所以\overrightarrow{DE}=(-1，1，0),\overrightarrow{CS}=(0，0，3),\overrightarrow{CD}=(1，1，0)，因为\overrightarrow{DE}\cdot \overrightarrow{CS}=0,\overrightarrow{DE}\cdot \overrightarrow{CD}=0,所以DE\perp CS,DE\perp CD,因为CS\cap CD=C，所以DE\perp 平面SCD.
(6分)$
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$![Image](https://cdntk.53zaixian.com/tq/1701760611602649090/img/compression/M_4a95be6b896f7eba603ce8fb131d31a7.jpg)
$(1)求证:DE\perp 平面SCD;$

Solution:
$证明：以C为原点建立空间直角坐标系，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1701760610528907267/img/compression/M_c0fbb76361816b3d10910378af01aa06.jpg)$则C(0，0，0)，E(0，2，0),S(0，0，3),A\left(\frac{3}{2},0,0\right)，D(1，1，0),B(0，3，0)，所以\overrightarrow{DE}=(-1，1，0),\overrightarrow{CS}=(0，0，3),\overrightarrow{CD}=(1，1，0)，因为\overrightarrow{DE}\cdot \overrightarrow{CS}=0,\overrightarrow{DE}\cdot \overrightarrow{CD}=0,所以DE\perp CS,DE\perp CD,因为CS\cap CD=C，所以DE\perp 平面SCD.
(6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$<img_3120>
$(2)求二面角A-SD-C的余弦值;$

Solution:
$由(1)知平面SCD的一个法向量为\overrightarrow{DE}=(-1，1，0),设平面ASD的法向量为n=(x,y，z)，\overrightarrow{AS}=\left(-\frac{3}{2},0,3\right)，\overrightarrow{SB}=\left(0,3,-3\right)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AS}=-\frac{3}{2}x+3z=0,\\ \boldsymbol{n}\cdot \overrightarrow{SB}=3y-3z=0,\end{matrix}\right.不妨取x=2，则n=(2，1，1).(9分)所以\\cos  \langle n，\overrightarrow{DE}\rangle =\frac{-2+1}{\sqrt{2}\times \sqrt{6}}=-\frac{\sqrt{3}}{6}，易知二面角A-SD-C的平面角为锐角，所以二面角A-SD-C的余弦值为\frac{\sqrt{3}}{6}.(12分)$

Final answer: $\frac{\sqrt{3}}{6}$
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$![Image](https://cdntk.53zaixian.com/tq/1701760611602649090/img/compression/M_4a95be6b896f7eba603ce8fb131d31a7.jpg)
$(2)求二面角A-SD-C的余弦值;$

Solution:
$由(1)知平面SCD的一个法向量为\overrightarrow{DE}=(-1，1，0),设平面ASD的法向量为n=(x,y，z)，\overrightarrow{AS}=\left(-\frac{3}{2},0,3\right)，\overrightarrow{SB}=\left(0,3,-3\right)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{AS}=-\frac{3}{2}x+3z=0,\\ \boldsymbol{n}\cdot \overrightarrow{SB}=3y-3z=0,\end{matrix}\right.不妨取x=2，则n=(2，1，1).(9分)所以\\cos  \langle n，\overrightarrow{DE}\rangle =\frac{-2+1}{\sqrt{2}\times \sqrt{6}}=-\frac{\sqrt{3}}{6}，易知二面角A-SD-C的平面角为锐角，所以二面角A-SD-C的余弦值为\frac{\sqrt{3}}{6}.(12分)$

Final answer: $\frac{\sqrt{3}}{6}$
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$![Image](https://cdntk.53zaixian.com/tq/1701760611602649090/img/compression/M_4a95be6b896f7eba603ce8fb131d31a7.jpg)
$(3)求点A到平面SCD的距离.$

Solution:
$因为平面SCD的一个法向量为\overrightarrow{DE}=(-1，1，0),\overrightarrow{CA}=\left(\frac{3}{2},0,0\right)，所以A到平面SCD的距离d=\left|\frac{\overrightarrow{CA}\cdot \overrightarrow{DE}}{|\overrightarrow{DE}|}\right|=\frac{\frac{3}{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{4}.(15分)$

Final answer: $\frac{3\sqrt{2}}{4}$
Subfield: Solid Geometry


Question: $如图,在三棱锥S-ABC中，SC\perp 平面ABC,SC=3,AC\perp BC,CE=2EB=2,AC=\frac{3}{2}，CD=ED.$<img_3120>
$(3)求点A到平面SCD的距离.$

Solution:
$因为平面SCD的一个法向量为\overrightarrow{DE}=(-1，1，0),\overrightarrow{CA}=\left(\frac{3}{2},0,0\right)，所以A到平面SCD的距离d=\left|\frac{\overrightarrow{CA}\cdot \overrightarrow{DE}}{|\overrightarrow{DE}|}\right|=\frac{\frac{3}{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{4}.(15分)$

Final answer: $\frac{3\sqrt{2}}{4}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$<img_2833>
$(1)证明：EF\perp BC；$

Solution:
$证明：连接A_{1}E，\because A_{1}A=A_{1}C，E是AC的中点，\therefore A_{1}E\perp AC，又平面A_{1}ACC_{1}\perp 平面ABC，A_{1}E\subset 平面A_{1}ACC_{1}，平面A_{1}ACC_{1}\cap 平面ABC=AC，\therefore A_{1}E\perp 平面ABC，\because BC\subset 平面ABC，\therefore A_{1}E\perp BC，\because A_{1}F\parallel AB，\angle ABC=90^\circ ，\therefore BC\perp A_{1}F，\because A_{1}F\cap A_{1}E=A_{1}，\therefore BC\perp 平面A_{1}EF，\because EF\subset 平面A_{1}EF,\therefore EF\perp BC.$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760612722528258/img/compression/M_8602c09481a1a460a090925a50299694.jpg)
$(1)证明：EF\perp BC；$

Solution:
$证明：连接A_{1}E，\because A_{1}A=A_{1}C，E是AC的中点，\therefore A_{1}E\perp AC，又平面A_{1}ACC_{1}\perp 平面ABC，A_{1}E\subset 平面A_{1}ACC_{1}，平面A_{1}ACC_{1}\cap 平面ABC=AC，\therefore A_{1}E\perp 平面ABC，\because BC\subset 平面ABC，\therefore A_{1}E\perp BC，\because A_{1}F\parallel AB，\angle ABC=90^\circ ，\therefore BC\perp A_{1}F，\because A_{1}F\cap A_{1}E=A_{1}，\therefore BC\perp 平面A_{1}EF，\because EF\subset 平面A_{1}EF,\therefore EF\perp BC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760612722528258/img/compression/M_8602c09481a1a460a090925a50299694.jpg)
$(2)求直线EF与平面A_{1}BC所成角的余弦值；$

Solution:
$如图，以E为原点，在平面ABC中，过E作AC的垂线为x轴，EC，EA_{1}所在直线分别为y，z轴，建立空间直角坐标系，E(0，0，0)，$![Image](https://cdntk.53zaixian.com/tq/1701760612965797889/img/compression/M_63917c042b8afbe84aca383574b1664d.jpg)$
设AC=4，易得A_{1}(0，0，2\sqrt{3})，B(\sqrt{3},1,0)，B_{1}(\sqrt{3},3,2\sqrt{3})，F\left(\frac{\sqrt{3}}{2},\frac{3}{2},2\sqrt{3}\right)，C(0，2，0)，则\overrightarrow{EF}=\left(\frac{\sqrt{3}}{2},\frac{3}{2},2\sqrt{3}\right)，\overrightarrow{BC}=(-\sqrt{3},1,0)，\overrightarrow{A_1C}=(0，2，-2\sqrt{3})，设平面A_{1}BC的法向量为n=(x，y，z)，则\left\{\begin{matrix}\overrightarrow{BC}\cdot \boldsymbol{n}=-\sqrt{3}x+y=0,\\ \overrightarrow{A_1C}\cdot \boldsymbol{n}=2y-2\sqrt{3}z=0,\end{matrix}\right.取x=1，得n=(1,\sqrt{3},1)，设直线EF与平面A_{1}BC所成角为\theta ，则\\sin   \theta =\frac{|\overrightarrow{EF}\cdot \boldsymbol{n}|}{|\overrightarrow{EF}|\cdot |\boldsymbol{n}|}=\frac{4}{5}，\therefore 直线EF与平面A_{1}BC所成角的余弦值为\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$<img_2833>
$(2)求直线EF与平面A_{1}BC所成角的余弦值；$

Solution:
$如图，以E为原点，在平面ABC中，过E作AC的垂线为x轴，EC，EA_{1}所在直线分别为y，z轴，建立空间直角坐标系，E(0，0，0)，$<img_1760>$
设AC=4，易得A_{1}(0，0，2\sqrt{3})，B(\sqrt{3},1,0)，B_{1}(\sqrt{3},3,2\sqrt{3})，F\left(\frac{\sqrt{3}}{2},\frac{3}{2},2\sqrt{3}\right)，C(0，2，0)，则\overrightarrow{EF}=\left(\frac{\sqrt{3}}{2},\frac{3}{2},2\sqrt{3}\right)，\overrightarrow{BC}=(-\sqrt{3},1,0)，\overrightarrow{A_1C}=(0，2，-2\sqrt{3})，设平面A_{1}BC的法向量为n=(x，y，z)，则\left\{\begin{matrix}\overrightarrow{BC}\cdot \boldsymbol{n}=-\sqrt{3}x+y=0,\\ \overrightarrow{A_1C}\cdot \boldsymbol{n}=2y-2\sqrt{3}z=0,\end{matrix}\right.取x=1，得n=(1,\sqrt{3},1)，设直线EF与平面A_{1}BC所成角为\theta ，则\\sin   \theta =\frac{|\overrightarrow{EF}\cdot \boldsymbol{n}|}{|\overrightarrow{EF}|\cdot |\boldsymbol{n}|}=\frac{4}{5}，\therefore 直线EF与平面A_{1}BC所成角的余弦值为\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760612722528258/img/compression/M_8602c09481a1a460a090925a50299694.jpg)
$(3)求二面角A-A_{1}C-B的正弦值.$

Solution:
$易知平面AA_{1}C的一个法向量为m=(1,0,0),则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{\sqrt{5}}{5},\therefore 二面角A-A_{1}C-B的正弦值为\sqrt{1-\left(\frac{\sqrt{5}}{5}\right)^2}=\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $如图，已知三棱柱ABC-A_{1}B_{1}C_{1}，平面A_{1}ACC_{1}\perp 平面ABC，\angle ABC=90^\circ ,\angle BAC=30^\circ ，A_{1}A=A_{1}C=AC，E，F分别是AC，A_{1}B_{1}的中点.$<img_2833>
$(3)求二面角A-A_{1}C-B的正弦值.$

Solution:
$易知平面AA_{1}C的一个法向量为m=(1,0,0),则\\cos  \langle m,n\rangle =\frac{\boldsymbol{m}\cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{\sqrt{5}}{5},\therefore 二面角A-A_{1}C-B的正弦值为\sqrt{1-\left(\frac{\sqrt{5}}{5}\right)^2}=\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$<img_2636>
$(1)证明:AC\perp SB;$

Solution:
$证明:取AC的中点D，连接BD，SD，$<img_2388>$\because SA=SC，\therefore SD\perp AC.同理BD\perp AC.又\because BD\cap SD=D,BD，SD\subset 平面SDB，\therefore AC\perp 平面SDB，\because SB\subset 平面SBD，\therefore AC\perp SB.$
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760614102454274/img/compression/M_79693340dded609bff296a7f7ffa5f6d.jpg)
$(1)证明:AC\perp SB;$

Solution:
$证明:取AC的中点D，连接BD，SD，$![Image](https://cdntk.53zaixian.com/tq/1701760613544611842/img/compression/M_fc6275bca0092d3eccef423ff4b6757e.jpg)$\because SA=SC，\therefore SD\perp AC.同理BD\perp AC.又\because BD\cap SD=D,BD，SD\subset 平面SDB，\therefore AC\perp 平面SDB，\because SB\subset 平面SBD，\therefore AC\perp SB.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760614102454274/img/compression/M_79693340dded609bff296a7f7ffa5f6d.jpg)
$(2)求二面角N-CM-B的正切值；$

Solution:
$由(1)知SD\perp AC，又\because 平面SAC\perp 平面ABC，AC为交线，SD\subset 平面SAC，\therefore SD\perp 平面ABC.\because BD\subset 平面ABC，\therefore SD\perp DB.在Rt\triangle SDB中，过点N作NP\parallel SD交BD于点P，\because N为SB的中点，\therefore P为BD的中点，且NP\perp 平面ABC.过点P作PQ\perp MC，垂足为Q，连接QN，易知QN\perp MC，\therefore \angle NQP为二面角N-CM-B的平面角.设BD与MC的交点为K，则K为\triangle ABC的重心，\because CM\perp AB，CM\perp PQ，\therefore PQ\parallel MB，故\triangle KPQ∽\triangle KBM.\therefore PQ:BM=PK:BK，\because KB=\frac{2}{3}BD=\frac{2}{3}\times \frac{\sqrt{3}}{2}AB=\frac{4\sqrt{3}}{3}，PK=\frac{1}{2}DB-\frac{1}{3}DB=\frac{1}{6}BD=\frac{\sqrt{3}}{3}，BM=2，\therefore PQ=\frac{BM\cdot PK}{BK}=\frac{1}{2}.\because NP=\frac{1}{2}SD=\frac{1}{2} \sqrt{SA^2-AD^2}=\sqrt{2}，\therefore \\tan  \angle NQP=\frac{NP}{PQ}=2\sqrt{2}，即二面角N-CM-B的正切值为2\sqrt{2}.$

Final answer: $2\sqrt{2}$
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$<img_2636>
$(2)求二面角N-CM-B的正切值；$

Solution:
$由(1)知SD\perp AC，又\because 平面SAC\perp 平面ABC，AC为交线，SD\subset 平面SAC，\therefore SD\perp 平面ABC.\because BD\subset 平面ABC，\therefore SD\perp DB.在Rt\triangle SDB中，过点N作NP\parallel SD交BD于点P，\because N为SB的中点，\therefore P为BD的中点，且NP\perp 平面ABC.过点P作PQ\perp MC，垂足为Q，连接QN，易知QN\perp MC，\therefore \angle NQP为二面角N-CM-B的平面角.设BD与MC的交点为K，则K为\triangle ABC的重心，\because CM\perp AB，CM\perp PQ，\therefore PQ\parallel MB，故\triangle KPQ∽\triangle KBM.\therefore PQ:BM=PK:BK，\because KB=\frac{2}{3}BD=\frac{2}{3}\times \frac{\sqrt{3}}{2}AB=\frac{4\sqrt{3}}{3}，PK=\frac{1}{2}DB-\frac{1}{3}DB=\frac{1}{6}BD=\frac{\sqrt{3}}{3}，BM=2，\therefore PQ=\frac{BM\cdot PK}{BK}=\frac{1}{2}.\because NP=\frac{1}{2}SD=\frac{1}{2} \sqrt{SA^2-AD^2}=\sqrt{2}，\therefore \\tan  \angle NQP=\frac{NP}{PQ}=2\sqrt{2}，即二面角N-CM-B的正切值为2\sqrt{2}.$

Final answer: $2\sqrt{2}$
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$![Image](https://cdntk.53zaixian.com/tq/1701760614102454274/img/compression/M_79693340dded609bff296a7f7ffa5f6d.jpg)
$(3)求点B到平面CMN的距离.$

Solution:
$解法一:设点B到平面CMN的距离为h.\because 在\triangle NCM中，NQ\perp CM，NQ=\sqrt{PQ^2+NP^2}=\frac{3}{2}，CM=2\sqrt{3}，\therefore S_{\triangle NCM}=\frac{1}{2}\times 2\sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{2}，S_{\triangle BCM}=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3},\therefore V_{N-BCM}=\frac{1}{3}S_{\triangle BCM}\cdot NP=\frac{1}{3}\times 2\sqrt{3}\times \sqrt{2}=\frac{2\sqrt{6}}{3}，V_{B-NCM}=\frac{1}{3}S_{\triangle NCM}\cdot h=\frac{\sqrt{3}}{2}h，\therefore \frac{\sqrt{3}}{2}h=\frac{2\sqrt{6}}{3}\Rightarrow h=\frac{4\sqrt{2}}{3}，即点B到平面CMN的距离为\frac{4\sqrt{2}}{3}.解法二:在Rt\triangle NPQ中，过P作PO\perp NQ，O为垂足，则可得PO\perp 平面CMN，又\because PO=\frac{NP\cdot PQ}{NQ}=\frac{\sqrt{2}}{3}，\therefore 点P到平面CMN的距离为\frac{\sqrt{2}}{3}.又\because BK=4PK，\therefore 点B到平面CMN的距离为\frac{4\sqrt{2}}{3}.$

Final answer: $\frac{4\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $在三棱锥S-ABC中，\triangle ABC是边长为4的正三角形，平面SAC\perp 平面ABC，SA=SC=2\sqrt{3}，M、N分别为AB、SB的中点.$<img_2636>
$(3)求点B到平面CMN的距离.$

Solution:
$解法一:设点B到平面CMN的距离为h.\because 在\triangle NCM中，NQ\perp CM，NQ=\sqrt{PQ^2+NP^2}=\frac{3}{2}，CM=2\sqrt{3}，\therefore S_{\triangle NCM}=\frac{1}{2}\times 2\sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{2}，S_{\triangle BCM}=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3},\therefore V_{N-BCM}=\frac{1}{3}S_{\triangle BCM}\cdot NP=\frac{1}{3}\times 2\sqrt{3}\times \sqrt{2}=\frac{2\sqrt{6}}{3}，V_{B-NCM}=\frac{1}{3}S_{\triangle NCM}\cdot h=\frac{\sqrt{3}}{2}h，\therefore \frac{\sqrt{3}}{2}h=\frac{2\sqrt{6}}{3}\Rightarrow h=\frac{4\sqrt{2}}{3}，即点B到平面CMN的距离为\frac{4\sqrt{2}}{3}.解法二:在Rt\triangle NPQ中，过P作PO\perp NQ，O为垂足，则可得PO\perp 平面CMN，又\because PO=\frac{NP\cdot PQ}{NQ}=\frac{\sqrt{2}}{3}，\therefore 点P到平面CMN的距离为\frac{\sqrt{2}}{3}.又\because BK=4PK，\therefore 点B到平面CMN的距离为\frac{4\sqrt{2}}{3}.$

Final answer: $\frac{4\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760614559633410/img/compression/M_f03a87e835f56775a1b1fbf4eed2e193.jpg)
$(1)求证:FG\perp AC；$

Solution:
$证明：连接BE，因为E，F分别为AC，A_{1}C_{1}的中点，所以EF\parallel CC_{1}，又BG\parallel CC_{1}，所以EF\parallel BG，所以E、F、B、G四点共面，易知EF\perp 平面ABC，因为AC\subset 平面ABC，所以EF\perp AC，又因为BE\perp AC，且EF\cap BE=E，所以AC\perp 平面EFGB，因为FG\subset 平面EFGB，所以FG\perp AC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$<img_2794>
$(1)求证:FG\perp AC；$

Solution:
$证明：连接BE，因为E，F分别为AC，A_{1}C_{1}的中点，所以EF\parallel CC_{1}，又BG\parallel CC_{1}，所以EF\parallel BG，所以E、F、B、G四点共面，易知EF\perp 平面ABC，因为AC\subset 平面ABC，所以EF\perp AC，又因为BE\perp AC，且EF\cap BE=E，所以AC\perp 平面EFGB，因为FG\subset 平面EFGB，所以FG\perp AC.$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$<img_2794>
$(2)求直线FG与平面A_{1}B_{1}C_{1}所成角的大小;$

Solution:
$以E为原点，建立如图所示的空间直角坐标系E-xyz，$<img_2669>$则E(0，0，0)，F(0，0，2)，G(0，\sqrt{3}，1)，所以\overrightarrow{FG}=(0，\sqrt{3}，-1)，易知\overrightarrow{EF}=(0，0，2)为平面A_{1}B_{1}C_{1}的一个法向量，设直线FG与平面A_{1}B_{1}C_{1}所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{FG},\overrightarrow{EF}\rangle |=\frac{|\overrightarrow{FG}\cdot \overrightarrow{EF}|}{|\overrightarrow{FG}|\cdot |\overrightarrow{EF}|}=\frac{2}{\sqrt{3+1}\times 2}=\frac{1}{2}，因为\theta \in \left[0,\frac{\pi }{2}\right]，所以\theta =\frac{\pi }{6}.$

Final answer: $\frac{\pi }{6}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC-A_{1}B_{1}C_{1}中，D为棱AA_{1}上的点，E，F，G分别为AC，A_{1}C_{1}，BB_{1}的中点，AC=AA_{1}=2.$![Image](https://cdntk.53zaixian.com/tq/1701760614559633410/img/compression/M_f03a87e835f56775a1b1fbf4eed2e193.jpg)
$(2)求直线FG与平面A_{1}B_{1}C_{1}所成角的大小;$

Solution:
$以E为原点，建立如图所示的空间直角坐标系E-xyz，$![Image](https://cdntk.53zaixian.com/tq/1701760614836457474/img/compression/M_8a08a4df1aa97dbe50aff4bf39ad73ec.jpg)$则E(0，0，0)，F(0，0，2)，G(0，\sqrt{3}，1)，所以\overrightarrow{FG}=(0，\sqrt{3}，-1)，易知\overrightarrow{EF}=(0，0，2)为平面A_{1}B_{1}C_{1}的一个法向量，设直线FG与平面A_{1}B_{1}C_{1}所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{FG},\overrightarrow{EF}\rangle |=\frac{|\overrightarrow{FG}\cdot \overrightarrow{EF}|}{|\overrightarrow{FG}|\cdot |\overrightarrow{EF}|}=\frac{2}{\sqrt{3+1}\times 2}=\frac{1}{2}，因为\theta \in \left[0,\frac{\pi }{2}\right]，所以\theta =\frac{\pi }{6}.$

Final answer: $\frac{\pi }{6}$