《2025年名校课堂九年级数学下册人教版》

13. (7分)计算:$\tan45^{\circ}\sin^{2}60^{\circ}-4\sin30^{\circ}\cos45^{\circ}+\sqrt{6}\tan30^{\circ}$.

答案： 解：原式$=1\times(\frac{\sqrt{3}}{2})^{2}-4\times\frac{1}{2}\times\frac{\sqrt{2}}{2}+\sqrt{6}\times\frac{\sqrt{3}}{3}=\frac{3}{4}-\sqrt{2}+\sqrt{2}=\frac{3}{4}$

14. (10分)在$Rt\triangle ABC$中,$\angle C$为直角,$\angle A$,$\angle B$,$\angle C$所对边的长分别为$a$,$b$,$c$,根据下列条件解直角三角形:
(1)已知$c = 20$,$\angle A = 60^{\circ}$.
(2)已知$a=\sqrt{15}$,$b = \sqrt{5}$.

答案： 解：
(1) $\because\angle A = 60^{\circ}$，$\angle C$为直角，$\therefore\angle B = 90^{\circ}-60^{\circ}=30^{\circ}$。$\because c = 20$，$\angle B = 30^{\circ}$，$\therefore b=\frac{c}{2}=10$，$a=\sqrt{20^{2}-10^{2}} = 10\sqrt{3}$。
(2) $\because\angle C$为直角，$a=\sqrt{15}$，$b=\sqrt{5}$，$\therefore c=\sqrt{a^{2}+b^{2}}=\sqrt{15 + 5}=2\sqrt{5}$。$\because\sin B=\frac{b}{c}=\frac{\sqrt{5}}{2\sqrt{5}}=\frac{1}{2}$，$\therefore\angle B = 30^{\circ}$。$\therefore\angle A = 90^{\circ}-30^{\circ}=60^{\circ}$。

15. (11分)如图,在$\triangle ABC$中,$\angle C = 90^{\circ}$,点$D$在$BC$上,$BD = 4$,$AD = BC$,$\cos\angle ADC=\frac{3}{5}$.
(1)求$CD$的长.
(2)求$\tan B$的值.

答案： 解：
(1) 在$Rt\triangle ACD$中，$\cos\angle ADC=\frac{CD}{AD}=\frac{3}{5}$。$\therefore$设$CD = 3x$，$AD = BC = 5x$，根据勾股定理，得$AC=\sqrt{AD^{2}-CD^{2}} = 4x$。$\because BD = 4$，$\therefore5x - 3x = 4$，解得$x = 2$。$\therefore CD = 3x = 6$。
(2) $\because$在$Rt\triangle ABC$中，$AC = 4x = 8$，$BC = 5x = 10$，$\therefore\tan B=\frac{AC}{BC}=\frac{8}{10}=\frac{4}{5}$。

16. (12分)如图,$AB$为$\odot O$的直径,$C$,$F$为$\odot O$上两点,且点$C$为$\overset{\frown}{BF}$的中点,过点$C$作$AF$的垂线,交$AF$的延长线于点$E$,交$AB$的延长线于点$D$.
(1)求证:$DE$是$\odot O$的切线.
(2)若$\odot O$的半径为3,$\tan D=\frac{3}{4}$,求$AE$的长.

答案： 解：
(1) 证明：连接$OC$。$\because$点$C$为$\overset{\frown}{BF}$的中点，$\therefore\overset{\frown}{BC}=\overset{\frown}{CF}$。$\therefore\angle BAC=\angle FAC$。$\because OA = OC$，$\therefore\angle OCA=\angle BAC$。$\therefore\angle OCA=\angle FAC$。$\therefore OC// AE$。$\because AE\perp DE$，$\therefore OC\perp DE$。又$\because OC$是$\odot O$的半径，$\therefore DE$是$\odot O$的切线。
(2) 在$Rt\triangle OCD$中，$\because\tan D=\frac{OC}{CD}=\frac{3}{4}$，$OC = 3$，$\therefore CD = 4$。$\therefore OD=\sqrt{OC^{2}+CD^{2}} = 5$。$\therefore AD = OD + AO = 8$。在$Rt\triangle ADE$中，$\because\sin D=\frac{AE}{AD}=\frac{OC}{OD}=\frac{3}{5}$，$\therefore AE=\frac{24}{5}$。