《2026年启东中学作业本九年级数学下册苏科版宿迁专版》

9. (10 分)计算:(1)$2\sin30°+3\cos60°-4\tan45°$;
(2)$\frac{\cos30°}{1+\sin30°}+\tan^260°$.

答案： 9.解:
(1)原式$=2 × \frac{1}{2} + 3 × \frac{1}{2} - 4 × 1 = -\frac{3}{2}$．
(2)原式$=\frac{\sqrt{3}}{1+\frac{1}{2}} + (\sqrt{3})^2 = \frac{\sqrt{3}}{\frac{3}{2}} + 3 = \frac{\sqrt{3}}{3} + 3$．

10.(12 分)如图,在 $ Rt\triangle ABC$ 中,$\angle C=90°$,$AB=10$,$\cos\angle ABC=\frac{3}{5}$,$D$ 为 $AC$ 边上一点,且 $\angle DBC=30°$,求 $AD$ 的长.

答案： 10.解:$\because \cos \angle ABC = \frac{3}{5}$，$\angle C = 90°$，$\therefore \frac{BC}{AB} = \frac{3}{5}$．
$\because AB = 10$，$\therefore BC = 10 × \frac{3}{5} = 6$，
$\therefore AC = \sqrt{AB^2 - BC^2} = \sqrt{10^2 - 6^2} = 8$．
$\because \angle DBC = 30°$，$\therefore CD = BC · \tan 30° = 6 × \frac{\sqrt{3}}{3} = 2\sqrt{3}$，$\therefore AD = AC - CD = 8 - 2\sqrt{3}$．

11.(15 分)如图,在 $ Rt\triangle ABC$ 中,$\angle ACB=90°$,$D$ 是 $AB$ 的中点,连接 $CD$,过点 $B$ 作 $CD$ 的垂线,交 $CD$ 的延长线于点 $E$. 已知 $AC=30$,$cosA=\frac{3}{5}$.
求:(1)线段 $CD$ 的长;
(2)$\sin\angle DBE$ 的值.

答案： 11.解:
(1)$\because$在 Rt$\triangle ABC$中，$\angle ACB = 90°$，$AC = 30$，$\cos A = \frac{AC}{AB} = \frac{3}{5}$，$\therefore AB = 50$．
$\because D$是$AB$的中点，$\therefore CD = \frac{1}{2}AB = 25$．
(2)过点$C$作$CF \perp AB$于点$F$．
$\because$在 Rt$\triangle ABC$中，$\angle ACB = 90°$，$AC = 30$，$AB = 50$，$\therefore BC = \sqrt{AB^2 - AC^2} = \sqrt{50^2 - 30^2} = 40$．
$\because S_{\triangle ABC} = \frac{1}{2}AC · BC = \frac{1}{2}CF · AB$，
$\therefore CF = AC · BC ÷ AB = 24$，
$\therefore DF = \sqrt{CD^2 - CF^2} = \sqrt{25^2 - 24^2} = 7$，
$\therefore \sin \angle DCF = \frac{DF}{CD} = \frac{7}{25}$．
$\because \angle DCF + \angle CDF = \angle DBE + \angle BDE = 90°$，$\angle CDF = \angle BDE$，$\therefore \angle DBE = \angle DCF$，
$\therefore \sin \angle DBE = \sin \angle DCF = \frac{7}{25}$．

12.(15 分)如图,$AB$ 与 $\odot O$ 相切于点 $A$,半径 $OC // AB$,$BC$ 与 $\odot O$ 相交于点 $D$,连接 $AD$.
(1)求证:$\angle OCA=\angle ADC$;
(2)若 $AD=2$,$tanB=\frac{1}{3}$,求 $OC$ 的长.

答案：
12.
(1)证明:连接$OA$交$BC$于点$F$，如答图．
$\because AB$是$\odot O$的切线，$\therefore \angle OAB = 90°$．
$\because OC // AB$，$\therefore \angle AOC = \angle OAB = 90°$．
$\because OC = OA$，$\therefore \angle OCA = 45°$．
$\because \angle ADC = \frac{1}{2} \angle AOC = 45°$，$\therefore \angle OCA = \angle ADC$；
(2)解:过点$A$作$AE \perp BC$于点$E$，如答图．
$\because \angle ADE = 45°$，
$\therefore \triangle ADE$是等腰直角三角形，
$\therefore AE = DE = \frac{\sqrt{2}}{2}AD = \sqrt{2}$．
$\because \tan B = \frac{AE}{BE} = \frac{1}{3}$，
$\therefore BE = 3AE = 3\sqrt{2}$，
$\therefore AB = \sqrt{BE^2 + AE^2} = \sqrt{18 + 2} = 2\sqrt{5}$．
在 Rt$\triangle ABF$中，$\tan B = \frac{AF}{AB} = \frac{1}{3}$，
$\therefore AF = \frac{1}{3}AB = \frac{2\sqrt{5}}{3}$．
$\because OC // AB$，$\therefore \angle OCF = \angle B$，
$\therefore \tan \angle OCF = \frac{OF}{OC} = \frac{1}{3}$．
设$OC = r$，则$OF = OA - AF = r - \frac{2\sqrt{5}}{3}$，
$\therefore 3(r - \frac{2\sqrt{5}}{3}) = r$，解得$r = \sqrt{5}$，$\therefore OC = \sqrt{5}$．