《2025年1课3练江苏人民出版社九年级数学下册北师大版》

6 (2023·泰州中考)如图，堤坝AB长为10 m，坡度i为1∶0.75，底端A在地面上，堤坝与对面的山之间有一深沟，山顶D处立有高20 m的铁塔CD.小明欲测量山高DE，他在A处看到铁塔顶端C刚好在视线AB上，又在坝顶B处测得塔底D的仰角α为26°35′.求堤坝高及山高DE.(参考数据：sin 26°35′≈0.45，cos 26°35′≈0.89，tan 26°35′≈0.50，小明身高忽略不计，结果精确到1 m)

答案：
如图，过点$B$作$BH\perp AE$于点$H$.

$\because$堤坝的坡度$i$为$1:0.75$，$\therefore$设$BH = 4x\ m$，则$AH = 3x\ m$，$\therefore AB=\sqrt{AH^{2}+BH^{2}} = 5x\ m = 10\ m$，$\therefore x = 2$，$\therefore AH = 6\ m$，$BH = 8\ m$.过点$B$作$BF\perp CE$于点$F$，则$EF = BH = 8\ m$，$BF = EH$.设$DF = a\ m$.$\because\alpha = 26^{\circ}35'$，$\therefore BF=\frac{DF}{\tan26^{\circ}35'}\approx\frac{a}{0.5}= 2a(m)$，$\therefore AE=(6 + 2a)m$.$\because$坡度$i$为$1:0.75$，$\therefore CE:AE=(20 + a + 8):(6 + 2a)=1:0.75$，$\therefore a = 12$，$\therefore DF = 12\ m$，$\therefore DE = DF + EF = 12 + 8 = 20(m)$.故堤坝高$8\ m$，山高$DE$约为$20\ m$.

7 新情境 测量桥塔 (2024·天津中考)综合与实践活动中，要用测角仪测量天津海河上一座桥的桥塔AB的高度(如图(1)).某学习小组设计了一个方案：如图(2)，点C，D，E依次在同一条水平直线上，DE = 36 m，EC⊥AB，垂足为C.在D处测得桥塔顶部B的仰角(∠CDB)为45°，测得桥塔底部A的俯角(∠CDA)为6°，又在E处测得桥塔顶部B的仰角(∠CEB)为31°.
(1)求线段CD的长；(结果取整数)
(2)求桥塔AB的高度.(结果取整数)
(参考数据：tan 31°≈0.6，tan 6°≈0.1)

答案：
(1)设$CD = x$，$\because DE = 36\ m$，$\therefore CE = CD + DE=(x + 36)m$.$\because EC\perp AB$，$\therefore\angle BCE=\angle ACD = 90^{\circ}$.$\because\angle CDB = 45^{\circ}$，$\tan\angle CDB=\frac{BC}{CD}$，$\therefore BC = CD\cdot\tan\angle CDB=x\cdot\tan45^{\circ}=x\ m$.$\because\angle CEB = 31^{\circ}$，$\tan\angle CEB=\frac{BC}{CE}$，$\therefore BC = CE\cdot\tan\angle CEB=(x + 36)\cdot\tan31^{\circ}$，$\therefore x=(x + 36)\cdot\tan31^{\circ}$，解得$x=\frac{36\times\tan31^{\circ}}{1 - \tan31^{\circ}}\approx\frac{36\times0.6}{1 - 0.6}= 54$.故线段$CD$的长约为$54\ m$.
(2)$\because\angle CDA = 6^{\circ}$，$\tan\angle CDA=\frac{AC}{CD}$，$\therefore AC = CD\cdot\tan\angle CDA = 54\times\tan6^{\circ}\approx54\times0.1 = 5.4(m)$.$\therefore AB = AC + BC = 5.4 + 54\approx59(m)$.故桥塔$AB$的高度约为$59\ m$.