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23. (8分)(营口中考)为了丰富学生的文化生活，学校利用假期组织学生到素质教育基地A和科技智能馆B参观学习. 如图，学生从学校出发，走到C处时，发现A处位于C处的北偏西25°方向上，B处位于C处的北偏西55°方向上，老师将学生分成甲、乙两组，甲组前往A处，乙组前往B处. 已知B处在A处的南偏西20°方向上，且这两处相距1 000米，求甲组学生比乙组学生大约多走多远(参考数据：$\sqrt{2}\approx1.41$，$\sqrt{6}\approx2.45$).
答案：
如图，过点$B$作$BE\perp AC$，垂足为$E$。由题意，得$\angle ACD = 25^{\circ}$，$\angle BCD = 55^{\circ}$，$\angle FAB = 20^{\circ}$，$AB = 1000$米，$CD// FA$，$\therefore\angle CAF=\angle ACD = 25^{\circ}$，$\therefore\angle BAC=\angle FAB+\angle CAF = 45^{\circ}$，$\angle ACB=\angle BCD-\angle ACD = 30^{\circ}$。在$Rt\triangle ABE$中，$\angle BAE = 45^{\circ}$，$\cos\angle BAE=\frac{AE}{AB}$，$\sin\angle BAE=\frac{BE}{AB}$，$\therefore AE = AB\cdot\cos45^{\circ}=1000\times\frac{\sqrt{2}}{2}=500\sqrt{2}$(米)，$BE = AB\cdot\sin45^{\circ}=1000\times\frac{\sqrt{2}}{2}=500\sqrt{2}$(米)。在$Rt\triangle BCE$中，$\angle BCE = 30^{\circ}$，$\therefore BC = 2BE = 2\times500\sqrt{2}=1000\sqrt{2}$(米)。$\because\tan\angle BCE=\frac{BE}{CE}$，$\therefore CE=\frac{BE}{\tan30^{\circ}}=\frac{500\sqrt{2}}{\frac{\sqrt{3}}{3}}=500\sqrt{6}$(米)，$\therefore AC = AE + CE=(500\sqrt{2}+500\sqrt{6})$米，$\therefore AC - BC = 500\sqrt{2}+500\sqrt{6}-1000\sqrt{2}=500\sqrt{6}-500\sqrt{2}\approx520$(米)，$\therefore$甲组学生比乙组学生大约多走520米
如图，过点$B$作$BE\perp AC$，垂足为$E$。由题意，得$\angle ACD = 25^{\circ}$，$\angle BCD = 55^{\circ}$，$\angle FAB = 20^{\circ}$，$AB = 1000$米，$CD// FA$，$\therefore\angle CAF=\angle ACD = 25^{\circ}$，$\therefore\angle BAC=\angle FAB+\angle CAF = 45^{\circ}$，$\angle ACB=\angle BCD-\angle ACD = 30^{\circ}$。在$Rt\triangle ABE$中，$\angle BAE = 45^{\circ}$，$\cos\angle BAE=\frac{AE}{AB}$，$\sin\angle BAE=\frac{BE}{AB}$，$\therefore AE = AB\cdot\cos45^{\circ}=1000\times\frac{\sqrt{2}}{2}=500\sqrt{2}$(米)，$BE = AB\cdot\sin45^{\circ}=1000\times\frac{\sqrt{2}}{2}=500\sqrt{2}$(米)。在$Rt\triangle BCE$中，$\angle BCE = 30^{\circ}$，$\therefore BC = 2BE = 2\times500\sqrt{2}=1000\sqrt{2}$(米)。$\because\tan\angle BCE=\frac{BE}{CE}$，$\therefore CE=\frac{BE}{\tan30^{\circ}}=\frac{500\sqrt{2}}{\frac{\sqrt{3}}{3}}=500\sqrt{6}$(米)，$\therefore AC = AE + CE=(500\sqrt{2}+500\sqrt{6})$米，$\therefore AC - BC = 500\sqrt{2}+500\sqrt{6}-1000\sqrt{2}=500\sqrt{6}-500\sqrt{2}\approx520$(米)，$\therefore$甲组学生比乙组学生大约多走520米
24. (12分)(苏州中考)如图①所示为某种可调节支撑架，BC为水平固定杆，竖直固定杆AB⊥BC，活动杆AD可绕点A旋转，CD为液压可伸缩支撑杆，已知AB = 10 cm，BC = 20 cm，AD = 50 cm.
(1)如图②，当活动杆AD处于水平状态时，求可伸缩支撑杆CD的长度；
(2)如图③，当活动杆AD绕点A由水平状态按逆时针方向旋转角度$\alpha$，且$\tan\alpha = \frac{3}{4}$($\alpha$为锐角)，求此时可伸缩支撑杆CD的长度.
(1)如图②，当活动杆AD处于水平状态时，求可伸缩支撑杆CD的长度；
(2)如图③，当活动杆AD绕点A由水平状态按逆时针方向旋转角度$\alpha$，且$\tan\alpha = \frac{3}{4}$($\alpha$为锐角)，求此时可伸缩支撑杆CD的长度.
答案：
(1) 如图①，过点$C$作$CE\perp AD$，垂足为$E$。易得四边形$ABCE$是矩形，$\therefore AB = CE = 10\ cm$，$BC = AE = 20\ cm$。$\because AD = 50\ cm$，$\therefore ED = AD - AE = 50 - 20 = 30(cm)$。在$Rt\triangle CED$中，由勾股定理，得$CD=\sqrt{CE^{2}+DE^{2}}=\sqrt{10^{2}+30^{2}}=10\sqrt{10}(cm)$，$\therefore$可伸缩支撑杆$CD$的长度为$10\sqrt{10}cm$
(2) 如图②，过点$D$作$DF\perp BC$交$BC$的延长线于点$F$，交$AD'$于点$G$。易得四边形$ABFG$是矩形，$\therefore AB = FG = 10\ cm$，$AG = BF$，$\angle AGF = 90^{\circ}$，$\therefore\angle AGD = 90^{\circ}$。在$Rt\triangle ADG$中，$\tan\alpha=\frac{DG}{AG}=\frac{3}{4}$，$\therefore$设$DG = 3x\ cm$，则$AG = 4x\ cm$。$\therefore$在$Rt\triangle ADG$中，由勾股定理，得$AD=\sqrt{AG^{2}+DG^{2}}=\sqrt{(4x)^{2}+(3x)^{2}}=5x(cm)$。$\because AD = 50\ cm$，$\therefore 5x = 50$，解得$x = 10$，$\therefore AG = 40\ cm$，$DG = 30\ cm$，$\therefore DF = DG + FG = 30 + 10 = 40(cm)$，$BF = AG = 40\ cm$。$\because BC = 20\ cm$，$\therefore CF = BF - BC = 40 - 20 = 20(cm)$。在$Rt\triangle CFD$中，由勾股定理，得$CD=\sqrt{CF^{2}+DF^{2}}=\sqrt{20^{2}+40^{2}}=20\sqrt{5}(cm)$，$\therefore$此时可伸缩支撑杆$CD$的长度为$20\sqrt{5}cm$
(1) 如图①，过点$C$作$CE\perp AD$，垂足为$E$。易得四边形$ABCE$是矩形，$\therefore AB = CE = 10\ cm$，$BC = AE = 20\ cm$。$\because AD = 50\ cm$，$\therefore ED = AD - AE = 50 - 20 = 30(cm)$。在$Rt\triangle CED$中，由勾股定理，得$CD=\sqrt{CE^{2}+DE^{2}}=\sqrt{10^{2}+30^{2}}=10\sqrt{10}(cm)$，$\therefore$可伸缩支撑杆$CD$的长度为$10\sqrt{10}cm$
(2) 如图②，过点$D$作$DF\perp BC$交$BC$的延长线于点$F$，交$AD'$于点$G$。易得四边形$ABFG$是矩形，$\therefore AB = FG = 10\ cm$，$AG = BF$，$\angle AGF = 90^{\circ}$，$\therefore\angle AGD = 90^{\circ}$。在$Rt\triangle ADG$中，$\tan\alpha=\frac{DG}{AG}=\frac{3}{4}$，$\therefore$设$DG = 3x\ cm$，则$AG = 4x\ cm$。$\therefore$在$Rt\triangle ADG$中，由勾股定理，得$AD=\sqrt{AG^{2}+DG^{2}}=\sqrt{(4x)^{2}+(3x)^{2}}=5x(cm)$。$\because AD = 50\ cm$，$\therefore 5x = 50$，解得$x = 10$，$\therefore AG = 40\ cm$，$DG = 30\ cm$，$\therefore DF = DG + FG = 30 + 10 = 40(cm)$，$BF = AG = 40\ cm$。$\because BC = 20\ cm$，$\therefore CF = BF - BC = 40 - 20 = 20(cm)$。在$Rt\triangle CFD$中，由勾股定理，得$CD=\sqrt{CF^{2}+DF^{2}}=\sqrt{20^{2}+40^{2}}=20\sqrt{5}(cm)$，$\therefore$此时可伸缩支撑杆$CD$的长度为$20\sqrt{5}cm$
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