《2026年通城学典全国中考试题分类精粹中考数学》

20. (2025·陕西)小涵和小宇想测量公园山坡上一个信号杆的高度。在征得家长同意后，他们带着工具前往测量。测量示意图如图所示，他们在坡面$FB$上的点$D$处安装测角仪$DE$，测得信号杆顶端$A$的仰角$\alpha$为$45^{\circ}$，$DE$与坡面的夹角$\beta$为$72.5^{\circ}$，又测得点$D$与信号杆底端$B$之间的距离$DB$为$22m$。已知$DE = 1.7m$，点$A$，$B$，$C$在同一条直线上，$AB$，$DE$均与水平线$FC$垂直。求信号杆的高$AB$（参考数据：$\sin72.5^{\circ}\approx0.95$，$\cos72.5^{\circ}\approx0.30$，$\tan72.5^{\circ}\approx3.17$）。

答案：
20. 如图，过点E作$EI \perp AC$于点I，过点D作$DH \perp AC$于点H。$\because AB$，$DE$均与水平线$FC$垂直，$\therefore DE // AC$。
$\therefore \angle DBH = \angle BDE = 72.5^{\circ}$。$\because DH \perp AC$，$\therefore \angle DHI = 90^{\circ}$。在$Rt \triangle DBH$中，$BD = 22 m$，$\sin 72.5^{\circ} = \frac{DH}{BD}$，则$DH = BD\sin 72.5^{\circ} \approx 22 × 0.95 = 20.9(m)$。在$Rt \triangle DBH$中，$BD = 22 m$，$\cos 72.5^{\circ} = \frac{BH}{BD}$，则$BH = BD\cos 72.5^{\circ} \approx 22 × 0.30 = 6.6(m)$。由题意，得$\angle EDH = \angle DHI = \angle HIE = 90^{\circ}$。$\therefore$四边形EDHI是矩形。$\therefore EI = DH = 20.9 m$。$\because \angle AIE = 45^{\circ}$，
$\angle AIE = 90^{\circ}$，$\therefore \angle EAI = 45^{\circ}$。$\therefore AI = EI = 20.9 m$。
$\therefore AB = AI + IH - BH = 20.9 + 1.7 - 6.6 = 16(m)$。$\therefore$信号杆的高AB约为$16 m$

21. (2025·长沙)如图，某景区内两条互相垂直的道路$a$，$b$交于点$M$，景点$A$，$B$在道路$a$上，景点$C$在道路$b$上。为了进一步提升景区品质，景区管委会在道路$b$上又开发了风景优美的景点$D$。经测得景点$C$位于景点$B$的北偏东$60^{\circ}$方向上，位于景点$A$的北偏东$30^{\circ}$方向上，景点$B$位于景点$D$的南偏西$45^{\circ}$方向上。已知$AB = 800m$。求：
(1)$\angle ACB$的度数；
(2)景点$C$与景点$D$之间的距离（结果保留根号）。

答案：
21.
(1)如图，根据题意，得$\angle CBE = 60^{\circ}$，$\angle CAF = 30^{\circ}$，
$\angle BDM = 45^{\circ}$，$BM // AF // DM$。$\therefore \angle BCM = \angle CBE = 60^{\circ}$，$\angle ACM = \angle CAF = 30^{\circ}$。$\therefore \angle ACB = \angle BCM - \angle ACM = 60^{\circ} - 30^{\circ} = 30^{\circ}$
(2) $\because \angle CBE = 60^{\circ}$，
$\therefore \angle CBM = 90^{\circ} - \angle CBE = 90^{\circ} - 60^{\circ} = 30^{\circ}$。由
(1)，得
$\angle ACB = 30^{\circ}$。$\therefore \angle ABC = \angle ACB = 30^{\circ}$。又$\because AB = 800 m$，
$\therefore AC = AB = 800 m$。在$Rt \triangle ACM$中，$\sin \angle ACM = \frac{AM}{AC}$，
$\cos \angle ACM = \frac{CM}{AC}$，$\therefore AM = AC · \sin \angle ACM = 800 × \sin 30^{\circ} = 800 × \frac{1}{2} = 400(m)$，$CM = AC · \cos \angle ACM = 800 × \cos 30^{\circ} = 800 × \frac{\sqrt{3}}{2} = 400\sqrt{3}(m)$。$\therefore BM = AB + AM = 800 + 400 = 1200(m)$。$\because \angle BDM = 45^{\circ}$，$BM \perp DM$，
$\therefore \angle BDM = \angle DBM = 45^{\circ}$。$\therefore DM = BM = 1200 m$。$\therefore DC = DM - CM = (1200 - 400\sqrt{3}) m$。$\therefore$景点C与景点D之间的距离为$(1200 - 400\sqrt{3}) m$