《2025年创新设计高考总复习数学浙江专版》

（2024·新高考Ⅱ卷）如图，平面四边形 $ ABCD $ 中，$ AB = 8 $，$ CD = 3 $，$ AD = 5\sqrt{3} $，$ \angle ADC = 90^{\circ} $，$ \angle BAD = 30^{\circ} $，点 $ E $，$ F $ 满足 $ \overrightarrow{AE} = \frac{2}{5}\overrightarrow{AD} $，$ \overrightarrow{AF} = \frac{1}{2}\overrightarrow{AB} $。将 $ \triangle AEF $ 沿 $ EF $ 翻折至 $ \triangle PEF $，使得 $ PC = 4\sqrt{3} $。
（1）证明：$ EF \perp PD $；
（2）求平面 $ PCD $ 与平面 $ PBF $ 所成的二面角的正弦值。$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$$\frac{8\sqrt{65}}{65}$

答案：

(1)证明　由题意知，$AE = \frac{2}{5}AD = 2\sqrt{3}$，$AF = \frac{1}{2}AB = 4$，又$\angle BAD = 30^{\circ}$，所以由余弦定理得$EF^{2} = AE^{2} + AF^{2} - 2AE \cdot AF \cdot \cos 30^{\circ} = 4$，故$EF = 2$。又$EF^{2} + AE^{2} = AF^{2}$，所以$EF \perp AE$。由$EF \perp AE$及翻折的性质知$EF \perp PE$，$EF \perp ED$，又$ED \cap PE = E$，$ED$，$PE \subset$平面$PED$，所以$EF \perp$平面$PED$。又$PD \subset$平面$PED$，所以$EF \perp PD$。
(2)解　法一　如图，连接$CE$，由题意知$DE = 3\sqrt{3}$，$CD = 3$，$\angle CDE = 90^{\circ}$，故$CE = \sqrt{DE^{2} + CD^{2}} = 6$。又$PE = AE = 2\sqrt{3}$，$PC = 4\sqrt{3}$，所以$PE^{2} + CE^{2} = PC^{2}$，故$PE \perp CE$。又$PE \perp EF$，$CE \cap EF = E$，$CE$，$EF \subset$平面$ABCD$，所以$PE \perp$平面$ABCD$。$EF$，$ED$，$PE$两两垂直，故以$E$为原点，$EF$，$ED$，$PE$所在直线分别为$x$，$y$，$z$轴建立空间直角坐标系，则$P(0,0,2\sqrt{3})$，$D(0,3\sqrt{3},0)$，$F(2,0,0)$，$A(0,-2\sqrt{3},0)$，$C(3,3\sqrt{3},0)$。 连接$PA$，则$\overrightarrow{PD} = (0,3\sqrt{3}, -2\sqrt{3})$，$\overrightarrow{DC} = (3,0,0)$，$\overrightarrow{AP} = (0,2\sqrt{3},2\sqrt{3})$，$\overrightarrow{AF} = (2,2\sqrt{3},0)$。设平面$PCD$的法向量$\boldsymbol{n}_{1} = (x_{1},y_{1},z_{1})$，则$\begin{cases} \boldsymbol{n}_{1} \cdot \overrightarrow{PD} = 3\sqrt{3}y_{1} - 2\sqrt{3}z_{1} = 0, \\ \boldsymbol{n}_{1} \cdot \overrightarrow{DC} = 3x_{1} = 0, \end{cases}$可取$\boldsymbol{n}_{1} = (0,2,3)$。设平面$PBF$即平面$PAF$的法向量$\boldsymbol{n}_{2} = (x_{2},y_{2},z_{2})$，则$\begin{cases} \boldsymbol{n}_{2} \cdot \overrightarrow{AP} = 2\sqrt{3}y_{2} + 2\sqrt{3}z_{2} = 0, \\ \boldsymbol{n}_{2} \cdot \overrightarrow{AF} = 2x_{2} + 2\sqrt{3}y_{2} = 0, \end{cases}$可取$\boldsymbol{n}_{2} = (\sqrt{3}, -1,1)$。$|\cos\langle\boldsymbol{n}_{1},\boldsymbol{n}_{2}\rangle| = \frac{|\boldsymbol{n}_{1} \cdot \boldsymbol{n}_{2}|}{|\boldsymbol{n}_{1}| \cdot |\boldsymbol{n}_{2}|} = \frac{1}{\sqrt{65}} = \frac{\sqrt{65}}{65}$。故平面$PCD$与平面$PBF$所成二面角的正弦值为$\sqrt{1 - \frac{1}{65}} = \frac{8\sqrt{65}}{65}$。法二　延长$FB$，$DC$交于点$G$，连接$PG$，点$G$，$P$是平面$PCD$与平面$PBF$的公共点，所以平面$PCD \cap$平面$PBF = PG$。通过计算得$CG = BG = BC = 2$，过点$F$作平面$PGD$的垂线，垂足为$M$，过$M$作$MN \perp PG$，交$PG$于$N$，连接$FN$，则$\angle FNM$是平面$PCD$与平面$PBF$所成的二面角的平面角。作$EQ \perp PD$于$Q$，由
(1)知$EF \perp$平面$PED$，$EF // CD$，所以$CD \perp EQ$，又$PD \perp EQ$，且$PD \cap CD = D$，$PD$，$CD \subset$平面$PGD$，所以$EQ \perp$平面$PGD$，$EQ$为点$E$到平面$PGD$的距离。由$EF // CD$，$CD \subset$平面$PGD$，$EF \not\subset$平面$PGD$，所以$EF //$平面$PGD$，则点$E$，$F$到平面$PGD$的距离相等，所以$FM = EQ$。在$Rt \triangle PED$中，$EQ = \frac{PE \cdot ED}{PD} = \frac{2\sqrt{3} \cdot 3\sqrt{3}}{\sqrt{39}} = \frac{18}{\sqrt{39}}$，所以$FM = EQ = \frac{18}{\sqrt{39}}$。在$Rt \triangle EDG$中，$EG^{2} = (3\sqrt{3})^{2} + 5^{2} = 52$，在$Rt \triangle PEG$中，$PG = \sqrt{(2\sqrt{3})^{2} + 5^{2}} = \sqrt{64} = 8$，在$Rt \triangle PFG$中，$\cos\angle PGF = \frac{6^{2} + 8^{2} - 4^{2}}{2 × 6 × 8} = \frac{7}{8}$，$\sin\angle PGF = \frac{\sqrt{15}}{8}$，因为$FN \perp PG$，所以$FN = FG \cdot \sin\angle PGF = 6 × \frac{\sqrt{15}}{8} = \frac{3\sqrt{15}}{4}$，所以平面$PCD$与平面$PBF$所成角的二面角的正弦值为$\sin\angle FNM = \frac{FM}{FN} = \frac{18}{\sqrt{39}} × \frac{4}{3\sqrt{15}} = \frac{8\sqrt{65}}{65}$。

典例
已知 $ \triangle ABC $ 与 $ \triangle BCD $ 所在平面垂直，且 $ AB = BC = BD $，$ \angle ABC = \angle DBC = 120^{\circ} $，则二面角 $ A - BD - C $ 的余弦值为____。
提醒：课后完成《一轮对点 91 练》B 本对点练 60

答案：
$-\frac{\sqrt{5}}{5}$　[过$A$作$AE \perp CB$的延长线于$E$，连接$DE$，$\because$平面$ABC \perp$平面$BCD$，平面$ABC \cap$平面$BCD = BC$，$AE \subset$平面$ABC$，$\therefore AE \perp$平面$BCD$，$\therefore E$点即为点$A$在平面$BCD$内的射影，$\therefore \triangle BDE$为$\triangle ABD$在平面$BCD$内的射影。设$AB = a$，则$AE = DE = AB \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{2}a$，$\therefore AD = \frac{\sqrt{6}}{2}a$，由余弦定理可得$\cos\angle ABD = \frac{1}{4}$，$\therefore \sin\angle ABD = \frac{\sqrt{15}}{4}$，$\therefore S_{\triangle ABD} = \frac{1}{2}a^{2} × \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{8}a^{2}$，又$BE = \frac{1}{2}a$，$\therefore S_{\triangle BDE} = \frac{1}{2} × \frac{\sqrt{3}}{2}a × \frac{1}{2}a = \frac{\sqrt{3}}{8}a^{2}$，设二面角$A - BD - E$为$\theta$，$\therefore \cos\theta = \frac{S_{\triangle BDE}}{S_{\triangle ABD}} = \frac{\sqrt{5}}{5}$。而二面角$A - BD - C$与二面角$A - BD - E$互补，$\therefore$二面角$A - BD - C$的余弦值为$-\frac{\sqrt{5}}{5}$。]