《2025年北大绿卡九年级数学下册人教版》

9. 如图，在△ABC中，AB = AC = 5，sin B = \frac{4}{5}，则BC的长是( )．

A. 3
B. 6
C. 8
D. 9

答案： B

10. 如图，在Rt△ABC中，∠C = 90°，AB = 10，sin A = \frac{2}{5}，求BC的长和tan B的值．

答案： 【解】$BC = AB\cdot\sin A = 10\times\frac{2}{5} = 4$.
在$Rt\triangle ABC$中，
$AC = \sqrt{AB^{2} - BC^{2}} = 2\sqrt{21}$，
$\therefore\tan B = \frac{AC}{BC} = \frac{2\sqrt{21}}{4} = \frac{\sqrt{21}}{2}$.

11. 如图，是用12个相似的直角三角形组成的图案. 若OA = 1，则OG =( )．

A. \frac{125\sqrt{5}}{64}
B. \frac{125}{64}
C. \frac{64}{27}
D. \frac{32\sqrt{3}}{27}

答案： C 【解析】$\because12$个相似的直角三角形，
$\therefore\angle BOA = \angle BOC = \cdots = \frac{360^{\circ}}{12} = 30^{\circ}$，
$\therefore\frac{OA}{OB} = \frac{OB}{OC} = \frac{OC}{OD} = \cdots = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
$\because OA = 1$，
$\therefore OB = \frac{2}{3}\sqrt{3} = 1\times\frac{2}{3}\sqrt{3}$，
$OC = \frac{4}{3} = 1\times(\frac{2}{3}\sqrt{3})^{2}$，
$OD = 1\times(\frac{2}{3}\sqrt{3})^{3} = \frac{8}{9}\sqrt{3}$，
$\cdots$，
$\therefore OG = 1\times(\frac{2}{3}\sqrt{3})^{6} = \frac{64}{27}$.
故选 C.

12. 某种风筝如图1所示，这个风筝骨架如图2所示，其中DE⊥AC于点B，AD = CD = m，AD与AC的夹角为α，则该骨架中AC的长度为( )．
A. mcos α B. mtan α C. 2mcos α D. 2mtan α

答案： C

13.【阅读理解】为计算tan 15°的三角函数值，我们可以构建Rt△ACB(如图)，使得∠C = 90°，∠ABC = 30°，延长CB使BD = AB，连接AD，可得到∠D = 15°，设AC = 1，所以tan 15° = \frac{AC}{CD} = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = 2 - \sqrt{3}. 类比这种方法，请你计算tan 67.5°的值为( )．


A. \sqrt{2} + 1
B. \sqrt{2} - 1
C. \sqrt{2}
D. \frac{1}{2}

答案：
A 【解析】在$Rt\triangle ACB$中，使得$\angle C = 90^{\circ}$，$\angle ABC = 45^{\circ}$，
延长$CB$到点$D$，使$BD = AB$，连接$AD$.

$\because\angle ABC$是$\triangle ABD$的一个外角，
$\therefore\angle ABC = \angle D + \angle BAD = 45^{\circ}$.
$\because BA = BD$，$\therefore\angle D = \angle BAD = 22.5^{\circ}$，
$\therefore\angle CAD = 90^{\circ} - \angle D = 67.5^{\circ}$.
设$AC = BC = 1$，则$AB = BD = \sqrt{2}AC = \sqrt{2}$，
$\therefore CD = BC + BD = \sqrt{2} + 1$.
在$Rt\triangle ACD$中，$\tan\angle CAD = \tan 67.5^{\circ} = \frac{CD}{AC} = \sqrt{2} + 1$.
故选 A.

14. 如图，在△ABC中，AD⊥BC，AE是BC边上的中线，AB = 10，AD = 6，tan ∠ACB = 1.
(1)求BC的长；
(2)求sin ∠DAE的值．

答案： 【解】
(1)$\because$在$Rt\triangle ABD$中，$AB = 10$，$AD = 6$，
$\therefore BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{10^{2} - 6^{2}} = 8$.
$\because$在$Rt\triangle ADC$中，$\tan\angle ACB = \frac{AD}{DC} = 1$，$\therefore DC = 6$，
$\therefore BC = BD + DC = 8 + 6 = 14$.
(2)$\because AE$是$BC$边上的中线，$\therefore BE = \frac{1}{2}BC = 7$，
$\therefore DE = BD - BE = 8 - 7 = 1$，
$\therefore AE = \sqrt{AD^{2} + DE^{2}} = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$，
$\therefore\sin\angle DAE = \frac{DE}{AE} = \frac{1}{\sqrt{37}} = \frac{\sqrt{37}}{37}$.

15. 如图，△ABC中，∠C = 90°，AD是角平分线，AC = 8，AD = \frac{16}{3}\sqrt{3}，求∠B的大小和BC，AB的长．

答案： 【解】在$\triangle ACD$中，$\because\angle ACD = 90^{\circ}$，
$\therefore CD = \sqrt{AD^{2} - AC^{2}} = \frac{8}{3}\sqrt{3}$，$\therefore\angle CAD = 30^{\circ}$.
又$AD$平分$\angle CAB$，
$\therefore\angle CAB = 60^{\circ}$，$\therefore\angle B = 30^{\circ}$.
又$\because\angle DAB = \angle DBA = 30^{\circ}$，$\therefore BD = AD = \frac{16}{3}\sqrt{3}$，
$\therefore BC = CD + BD = 8\sqrt{3}$，$\therefore AB = 2AC = 16$.