《2025年通城学典课时作业本九年级数学下册苏科版江苏专版》

6. 如图，在△ABC中，AB = AC，以AB为直径的⊙O交AC于点D，交BC于点E，直线EF⊥AC于点F，交AB的延长线于点H.
　(1) 求证：HF是⊙O的切线；
　(2) 当EB = 6，cos∠ABE = $\frac{1}{3}$时，求tan H的值.
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

答案：
(1) 连接$OE$、$AE$.$\because AB$为$\odot O$的直径，$\therefore \angle AEB = 90^{\circ}$，即$AE \perp BC$.$\because AB = AC$，$\therefore$易得$BE = CE$.$\because OB = OA$，$\therefore OE // AC$.又$\because HF \perp AC$，$\therefore OE \perp HF$.$\because OE$是$\odot O$的半径，$\therefore HF$是$\odot O$的切线。
(2) 过点$E$作$EG \perp AH$于点$G$.$\because$在$Rt\triangle BGE$中，$\cos \angle ABE = \frac{1}{3}$，$EB = 6$，$\therefore BG = EB \cdot \cos \angle ABE = 2$.$\therefore EG = \sqrt{EB^{2} - BG^{2}} = 4\sqrt{2}$.$\because$在$Rt\triangle BEA$中，$\cos \angle ABE = \frac{1}{3}$，$EB = 6$，$\therefore AB = \frac{EB}{\cos \angle ABE} = 18$.$\therefore OB = OE = \frac{1}{2}AB = 9$.$\therefore GO = OB - BG = 7$.$\because EG \perp AH$，$OE \perp HF$，$\therefore \angle H + \angle HEG = 90^{\circ}$，$\angle GEO + \angle HEG = 90^{\circ}$.$\therefore \angle H = \angle GEO$.$\therefore$在$Rt\triangle EGO$中，$\tan H = \tan \angle GEO = \frac{GO}{EG} = \frac{7}{4\sqrt{2}} = \frac{7\sqrt{2}}{8}$

7. 如图，AD是⊙O的直径，BD、BC都是弦，且BD = BC，经过点B作⊙O的切线，交AD的延长线于点E.
　(1) 求证：∠EBD = ∠CAB；
　(2) 若BC = $\sqrt{3}$，AC = 5，求sin∠CBA的值.
　　

答案：
(1) 连接$OB$.$\because BE$是$\odot O$的切线，$\therefore OB \perp BE$.$\therefore \angle OBD + \angle EBD = 90^{\circ}$.$\because AD$是$\odot O$的直径，$\therefore \angle ABD = 90^{\circ}$.$\therefore \angle ABO + \angle OBD = 90^{\circ}$.$\therefore \angle EBD = \angle ABO$.$\because OA = OB$，$\therefore \angle OAB = \angle ABO$.$\therefore \angle OAB = \angle EBD$.$\because BD = BC$，$\therefore \overset{\frown}{BD} = \overset{\frown}{BC}$.$\therefore \angle BAD = \angle CAB$.$\therefore \angle EBD = \angle CAB$。
(2) 连接$CD$，交$OB$于点$M$.$\because \overset{\frown}{BD} = \overset{\frown}{BC}$，$\therefore CM = MD$，$OB \perp CD$.又$\because OA = OD$，$\therefore OM$为$\triangle ACD$的中位线.$\therefore OM = \frac{1}{2}AC = \frac{5}{2}$.设$\odot O$的半径为$r$，则$BM = r - \frac{5}{2}$.$\therefore$在$Rt\triangle OMD$和$Rt\triangle BMD$中，由勾股定理，得$DM^{2} = OD^{2} - OM^{2} = BD^{2} - BM^{2}$.$\because BD = BC = \sqrt{3}$，$\therefore r^{2} - (\frac{5}{2})^{2} = (\sqrt{3})^{2} - (r - \frac{5}{2})^{2}$，解得$r = 3$（负值舍去）.$\therefore AD = 2r = 6$.$\because AD$是$\odot O$的直径，$\therefore \angle ACD = 90^{\circ}$.$\therefore$在$Rt\triangle ACD$中，$\sin \angle ADC = \frac{AC}{AD} = \frac{5}{6}$.$\because \overset{\frown}{AC} = \overset{\frown}{AC}$，$\therefore \angle CBA = \angle ADC$.$\therefore \sin \angle CBA = \frac{5}{6}$

8. （2023·武汉）如图，在四边形ABCD中，AB//CD，AD⊥AB，以点D为圆心、AD长为半径的弧恰好与BC相切，切点为E. 若$\frac{AB}{CD}$ = $\frac{1}{3}$，则sin C的值为_______.
　　　　　　　　　　　　

答案： $\frac{\sqrt{5}}{3}$

9. 如图，半径为$\sqrt{3}$的⊙O与边长为8的等边三角形ABC的两边AB、BC都相切，连接OC，则tan∠OCB的值为_______.
　　　　　　

答案： $\frac{\sqrt{3}}{5}$ 解析：设$\odot O$与边$BC$相切于点$D$，连接$OB$、$OD$.$\because \odot O$与等边三角形$ABC$的两边$AB$、$BC$都相切，$\therefore$易得$\angle OBC = \angle OBA = \frac{1}{2} \angle ABC = 30^{\circ}$，$\angle ODB = 90^{\circ}$.$\because$在$Rt\triangle OBD$中，$\tan \angle OBD = \frac{OD}{BD}$，$\therefore BD = \frac{OD}{\tan \angle OBD} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{3}} = 3$.$\therefore CD = BC - BD = 8 - 3 = 5$.$\therefore$在$Rt\triangle OCD$中，$\tan \angle OCB = \frac{OD}{CD} = \frac{\sqrt{3}}{5}$