《2025年阳光夺冠九年级数学下册》

18. (8分)如图，在$Rt\triangle ABC$中，$\angle C = 90^{\circ}$，$\angle A = 60^{\circ}$，$\angle A$，$\angle B$，$\angle C$的对边分别为$a$，$b$，$c$，$a + b = 3+\sqrt{3}$. 请你据此条件，解这个直角三角形.

答案： 解：由题意，易得$\angle B = 90^{\circ}-\angle A = 30^{\circ}$.
$\because \tan A=\frac{a}{b}=\sqrt{3}$，$\therefore a=\sqrt{3}b$.
又$\because a + b = 3+\sqrt{3}$，$\therefore a = 3$，$b=\sqrt{3}$，$\therefore c=\sqrt{a^{2}+b^{2}}=2\sqrt{3}$.

19. (9分)如图，$AD$是$\triangle ABC$的高，$\cos B=\frac{\sqrt{2}}{2}$，$\sin C=\frac{3}{5}$，$AC = 10$，求$\triangle ABC$的周长.

答案： $\triangle ABC$的周长为$24 + 6\sqrt{2}$.

20. (10分)将一副直角三角板按如图所示放置，点$C$在$FD$的延长线上，$AB// CF$，$\angle F = \angle ACB = 90^{\circ}$，$\angle E = 45^{\circ}$，$\angle A = 60^{\circ}$，$AC = 10$，试求$CD$的长.

答案： 解：过点$B$作$BM\perp FD$于点$M$.
在$\triangle ACB$中，$\angle ACB = 90^{\circ}$，$\angle A = 60^{\circ}$，$AC = 10$，
$\therefore \angle ABC = 30^{\circ}$，$BC = AC\cdot\tan60^{\circ}=10\sqrt{3}$.
$\because AB// CF$，
$\therefore BM = BC\cdot\sin30^{\circ}=10\sqrt{3}\times\frac{1}{2}=5\sqrt{3}$，$CM = BC\cdot\cos30^{\circ}=10\sqrt{3}\times\frac{\sqrt{3}}{2}=15$.
在$\triangle EFD$中，$\angle F = 90^{\circ}$，$\angle E = 45^{\circ}$，
$\therefore \angle EDF = 45^{\circ}$，$\therefore MD = BM = 5\sqrt{3}$，
$\therefore CD = CM - MD = 15 - 5\sqrt{3}$.