《2025年1课3练江苏人民出版社九年级数学下册北师大版》

15 如图，已知在△ABC中，CD⊥AB，垂足为D，AD = 2，BD = 6，$\tan B=\frac{2}{3}$，点E是边BC的中点.
(1)求边AC的长；
(2)求∠EAB的正弦值.
　　　　　　　　　　

答案：

(1)$\because CD\perp AB$,
$\therefore\triangle ACD,\triangle BCD$均为直角三角形.
在$Rt\triangle CDB$中,
$\because BD = 6,\tan B=\frac{CD}{BD}=\frac{2}{3},\therefore CD = 4$.
在$Rt\triangle CDA$中,
$AC=\sqrt{CD^{2}+AD^{2}}=\sqrt{4^{2}+2^{2}} = 2\sqrt{5}$.
(2)如图,过点$E$作$EF\perp AB$,垂足为$F$.

$\because CD\perp AB,EF\perp AB$,
$\therefore CD// EF$.
又点$E$是边$BC$的中点,
$\therefore EF$是$\triangle BCD$的中位线.
$\therefore DF = BF = 3,EF=\frac{1}{2}CD = 2$.
$\therefore AF = AD + DF = 5$.
在$Rt\triangle AEF$中,
$AE=\sqrt{AF^{2}+EF^{2}}=\sqrt{5^{2}+2^{2}}=\sqrt{29}$,
$\therefore\sin\angle EAB=\frac{EF}{AE}=\frac{2}{\sqrt{29}}=\frac{2\sqrt{29}}{29}$.

16 (2024·上海浦东新区期末)如图，已知在四边形ABCD中，AD//BC，∠ABC = 90°，对角线AC，BD相交于点O，AD = 2，AB = 3，BC = 4.
(1)求△BOC的面积；
(2)求∠ACD的正弦值.
　　　　　　　　　　　

答案：

(1)如图,过点$O$作$AB$的平行线,分别与$AD,BC$交于点$M,N$.
$\because AD// BC,MN// AB$,
$\therefore$四边形$ABNM$是平行四边形.
又$\angle ABC = 90^{\circ},\therefore$四边形$ABNM$是矩形,
$\therefore OM\perp AD,ON\perp BC,MN = AB$.
$\because AD// BC,\therefore\triangle AOD\sim\triangle COB$,
$\therefore\frac{OM}{ON}=\frac{AD}{BC}=\frac{2}{4}=\frac{1}{2}$.
又$MN = AB = 3,\therefore OM = 1,ON = 2$,
$\therefore S_{\triangle BOC}=\frac{1}{2}BC\cdot ON=\frac{1}{2}\times4\times2 = 4$.
(2)在$Rt\triangle ABC$中,$AC=\sqrt{3^{2}+4^{2}} = 5$.
如图,过点$D$作$BC$的垂线,垂足为$E$,过点$A$作$CD$垂线,垂足为$F$,
在$Rt\triangle CDE$中,$CD=\sqrt{2^{2}+3^{2}}=\sqrt{13}$.
$\because S_{\triangle ACD}=\frac{1}{2}\times2\times3=\frac{1}{2}\times\sqrt{13}\times AF$,
$\therefore AF=\frac{6\sqrt{13}}{13}$.
在$Rt\triangle CAF$中,$\sin\angle ACD=\frac{AF}{AC}=\frac{\frac{6\sqrt{13}}{13}}{5}=\frac{6\sqrt{13}}{65}$.

17 跨学科 光的折射 如图，光从空气斜射入水中，入射光线AB射到水池的水面点B后折射光线BD射到池底点D处，入射角∠ABM = 30°，折射角∠DBN = 22°；入射光线AC射到水池的水面点C后折射光线CE射到池底点E处，入射角∠ACM' = 60°，折射角∠ECN' = 40.5°.DE//BC，MN，M'N'为法线.入射光线AB，AC和折射光线BD，CE及法线MN，M'N'都在同一平面内，点A到直线BC的距离为6米.
(1)求BC的长；(结果保留根号)
(2)如果DE = 8.72米，求水池的深.(参考数据：$\sqrt{2}\approx1.41,\sqrt{3}\approx1.73,\sin 22^{\circ}\approx0.37,\cos 22^{\circ}\approx0.93,\tan 22^{\circ}\approx0.4,\sin 40.5^{\circ}\approx0.65,\cos 40.5^{\circ}\approx0.76,\tan 40.5^{\circ}\approx0.85$)
　　　　　　

答案：

(1)如图,过点$A$作$AF\perp BC$,交$CB$的延长线于点$F$,则$AF// MN// M'N'$,
$\therefore\angle ABM=\angle BAF,\angle ACM'=\angle CAF$.
$\because\angle ABM = 30^{\circ},\angle ACM' = 60^{\circ}$,
$\therefore\angle BAF = 30^{\circ},\angle CAF = 60^{\circ}$.
$\because AF = 6$米,$\therefore BF = AF\cdot\tan30^{\circ}=6\times\frac{\sqrt{3}}{3}=2\sqrt{3}$(米),
$\therefore CF = AF\cdot\tan60^{\circ}=6\times\sqrt{3}=6\sqrt{3}$(米),
$\therefore BC = CF - BF = 6\sqrt{3}-2\sqrt{3}=4\sqrt{3}$(米).
即$BC$的长为$4\sqrt{3}$米.

(2)设水池的深为$x$米,则$BN = CN' = x$米,
由题意,得$\angle DBN = 22^{\circ},\angle ECN' = 40.5^{\circ}$,
$DE = 8.72$米,
$\therefore DN = BN\cdot\tan22^{\circ}\approx0.4x$米,$N'E = CN'\cdot\tan40.5^{\circ}\approx0.85x$米.
$\because DN + DE = BC + N'E$,
$\therefore 0.4x + 8.72 = 4\sqrt{3}+0.85x$,解得$x\approx4$.
即水池的深约为$4$米.