《2026年1号卷中考试题精编九年级数学安徽专版》

19. 科技社团选择学校游泳池进行一次光的折射实验,如图,光线自点$B$处发出,经水面点$E$折射到池底点$A$处.已知$BE$与水平线的夹角$\alpha =36.9^{\circ}$,点$B$到水面的距离$BC=1.20\ m$.点$A$处水深为$1.20\ m$,到池壁的水平距离$AD=2.50\ m$.点$B$,$C$,$D$在同一条竖直线上,所有点都在同一竖直平面内,记入射角为$\beta$,折射角为$\gamma$,求$\dfrac{\sin\beta}{\sin\gamma}$的值.(精确到$0.1$)
参考数据:$\sin 36.9^{\circ}\approx 0.60$,$\cos 36.9^{\circ}\approx 0.80$,$\tan 36.9^{\circ}\approx 0.75$.

答案： 19. 解:如图,过点$E$作$EH\perp AD$,垂足为点$H$. 由题意可知,$\angle CEB = \alpha = 36.9^{\circ}$,$EH = 1.20(m)$, $CE=\frac{BC}{\tan 36.9^{\circ}}\approx\frac{1.20}{0.75}=1.60(m)$,(4分) $\therefore AH = AD - CE = 2.50 - 1.60 = 0.90(m)$, $\therefore AE=\sqrt{AH^2 + EH^2}=\sqrt{0.90^2 + 1.20^2}=1.50(m)$, $\therefore \sin \gamma =\frac{AH}{AE}=\frac{0.90}{1.50}=0.60$.(8分) 又$\because \sin \beta = \sin \angle CBE = \frac{CE}{BE}=\cos \angle CEB = \cos \alpha \approx 0.80$,$\therefore \frac{\sin \beta}{\sin \gamma}\approx\frac{0.80}{0.60}\approx1.3$.(10分)

20. 如图,$\odot O$是$\triangle ABC$的外接圆,$D$是直径$AB$上一点,$\angle ACD$的平分线交$AB$于点$E$,交$\odot O$于另一点$F$,$FA=FE$.
(1)求证:$CD\perp AB$.
(2)设$FM\perp AB$,垂足为点$M$,若$OM=OE=1$,求$AC$的长.

答案： 20. 解:
(1)证明:$\because FA = FE$,$\therefore \angle FAE = \angle AEF$. $\because \angle FAE$与$\angle BCE$都是$BF$所对的圆周角, $\therefore \angle FAE = \angle BCE$. 又$\because \angle AEF = \angle CEB$,$\therefore \angle CEB = \angle BCE$. $\because CE$平分$\angle ACD$,$\therefore \angle ACE = \angle DCE$. 又$\because AB$是直径,$\therefore \angle ACB = 90^{\circ}$, $\therefore \angle CEB + \angle DCE = \angle BCE + \angle ACE = \angle ACB = 90^{\circ}$, $\therefore \angle CDE = 90^{\circ}$,$\therefore CD\perp AB$.(5分)
(2)由
(1)知$\angle BEC = \angle BCE$,$\therefore BE = BC$. 又$\because AF = EF$,$FM\perp AB$, $\therefore MA = ME = 2$,$AE = 4$. $\therefore OA = OB = AE - OE = 3$, $\therefore BC = BE = OB - OE = 2$. 在$\triangle ABC$中,$AB = 6$,$BC = 2$,$\angle ACB = 90^{\circ}$, $\therefore AC=\sqrt{AB^2 - BC^2}=\sqrt{6^2 - 2^2}=4\sqrt{2}$.(10分)