《2026年启东中学作业本九年级数学下册苏科版宿迁专版》

10.(2024·浙江)如图,在$\bigtriangleup ABC$中,$AD\bot BC$,$AE$是$BC$边上的中线,$AB = 10$,$AD = 6$,$\tan\angle ACB = 1$.
求:(1)$BC$的长;
(2)$\sin\angle DAE$的值.

答案： 10.解：
(1)$\because AD \perp BC$，$AB = 10$，$AD = 6$，
$\therefore BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{10^{2} - 6^{2}} = 8$．
$\because \tan\angle ACB = 1$，$\angle ADC = 90^{\circ}$，$\therefore CD = AD = 6$，
$\therefore BC = BD + CD = 8 + 6 = 14$．
(2)$\because AE$是$BC$边上的中线，
$\therefore CE = \frac{1}{2}BC = 7$，$\therefore DE = CE - CD = 7 - 6 = 1$．
$\because AD \perp BC$，$\therefore AE = \sqrt{AD^{2} + DE^{2}} = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$，
$\therefore \sin\angle DAE = \frac{DE}{AE} = \frac{1}{\sqrt{37}} = \frac{\sqrt{37}}{37}$．

11. 在 Rt$\bigtriangleup ABC$中,$a$,$b$,$c$分别为$\angle A$,$\angle B$,$\angle C$的对边.$\angle C = 90°$,$\angle A - \angle B = 30°$,$a - b = 2\sqrt{3} - 2$,解这个直角三角形.

答案： 11.解：$\because \begin{cases} \angle A - \angle B = 30^{\circ} \\ \angle A + \angle B = 90^{\circ} \end{cases}$，$\therefore \begin{cases} \angle A = 60^{\circ} \\ \angle B = 30^{\circ} \end{cases}$．
$\because \tan A = \frac{a}{b}$，$\therefore a = b\tan A = b\tan 60^{\circ} = \sqrt{3}b$，
由$\begin{cases} a - b = 2\sqrt{3} - 2 \\ a = \sqrt{3}b \end{cases}$解得$\begin{cases} a = 2\sqrt{3} \\ b = 2 \end{cases}$，
$\therefore c = \sqrt{a^{2} + b^{2}} = \sqrt{(2\sqrt{3})^{2} + 2^{2}} = 4$．

12. 如图,在$\bigtriangleup ABC$,$\angle B = 45°$,$CD$是$AB$边上的中线,过点$D$作$DE\bot BC$,垂足为$E$,若$CD = 15$,$\cos\angle BCD = \frac{4}{5}$.
求:(1)$\bigtriangleup BCD$的面积;
(2)$\angle ACB$的正切值.

答案：
12.解：
(1)$\because \cos\angle BCD = \frac{4}{5}$，$DE \perp BC$，$\therefore \frac{EC}{CD} = \frac{4}{5}$．
$\because CD = 15$，$\therefore CE = 12$，$\therefore DE = \sqrt{DC^{2} - EC^{2}} = 9$．
$\because \angle B = 45^{\circ}$，$\therefore DE = BE = 9$，
$\therefore BC = BE + CE = 21$，
$\therefore S_{\triangle BCD} = \frac{1}{2}BC · DE = \frac{1}{2} × 21 × 9 = \frac{189}{2}$．
(2)过点$A$作$AF \perp BC$于点$F$，如答图，$\therefore DE // AF$．
$\because D$是$AB$的中点，$\therefore DE$是$\triangle ABF$的中位线，$\therefore AF = 2DE$，$BF = 2BE$．
由
(1)可知$DE = BE = 9$，$\therefore AF = BF = 18$，$\therefore CF = BC - BF = 3$，$\therefore \tan\angle ACB = \frac{AF}{CF} = 6$．

13. 如图,在$\bigtriangleup ABC$中,$\angle ACB = 90°$,$\angle CAB ＞ \angle B$,$CD$是斜边$AB$上的中线,过点$A$作$\angle CAE = \angle B$,交$BC$于点$E$,交$CD$于点$H$,且$AH = 2CH$.
(1)求$\sin B$的值;
(2)当$CD = \sqrt{5}$时,求$BE$的长.

答案： 13.解：
(1)$\because \angle ACB = 90^{\circ}$，$CD$是斜边$AB$上的中线，
$\therefore CD = BD$，$\therefore \angle B = \angle DCB$．
$\because \angle CAE = \angle B$，$\therefore \angle CAE = \angle DCB$．
$\because \angle ACD + \angle DCB = 90^{\circ}$，$\therefore \angle ACD + \angle CAE = 90^{\circ}$，
$\therefore \angle AHC = 180^{\circ} - (\angle ACD + \angle CAE) = 90^{\circ}$，
$\therefore AE \perp CD$．
$\because AH = 2CH$，
$\therefore AC = \sqrt{AH^{2} + CH^{2}} = \sqrt{(2CH)^{2} + CH^{2}} = \sqrt{5}CH$，
$\therefore \sin\angle CAE = \frac{CH}{AC} = \frac{CH}{\sqrt{5}CH} = \frac{\sqrt{5}}{5}$，
$\therefore \sin B = \frac{\sqrt{5}}{5}$．
(2)$\because \angle ACB = 90^{\circ}$，$CD$是斜边$AB$上的中线，$CD = \sqrt{5}$．
$\therefore AB = 2CD = 2\sqrt{5}$．
在$Rt\triangle ABC$中，$\because \sin B = \frac{AC}{AB} = \frac{\sqrt{5}}{5}$，$\therefore AC = 2$，
$\therefore BC = \sqrt{AB^{2} - AC^{2}} = \sqrt{(2\sqrt{5})^{2} - 2^{2}} = 4$．
$\because$在$Rt\triangle ACH$中，$\tan\angle CAE = \frac{CH}{AH} = \frac{1}{2}$，
$\therefore$在$Rt\triangle ACE$中，$\tan\angle CAE = \frac{CE}{AC} = \frac{1}{2}$，$\therefore CE = 1$，
$\therefore BE = BC - CE = 4 - 1 = 3$，$\therefore BE$的长为$3$．