1. 在$Rt△ ABC$中，$∠ C = 90°$，$a,b,c$分别是$∠ A$，$∠ B$，$∠ C$的对边.
(1) 若$a = 4\sqrt{3}$，$b = 4$，则$∠ A =$ ，$∠ B =$ ，$c =$ ；
(2) 若$c = 4$，$b = 2\sqrt{2}$，则$a =$ ，$∠ B =$ ，$∠ A =$ .

(1)$\because \tan A = \frac{a}{b} = \sqrt{3},\therefore ∠ A = 60°,\therefore ∠ B = 30°$. 由$a = 4\sqrt{3},b = 4$,得 $c = \sqrt{a^2 + b^2} = 8$.
(2)$\because c = 4, b = 2\sqrt{2}, \therefore \cos A = \frac{b}{c} = \frac{\sqrt{2}}{2}$,$\therefore ∠ A = ∠ B = 45°,\therefore a = b = 2\sqrt{2}$.

2. 在$\mathrm{Rt}△ ABC$中，$∠ C=90°$，$∠ A$，$∠ B$，$∠ C$所对的边分别为$a$，$b$，$c$，且$a=4\sqrt{15}$，$c=8\sqrt{5}$，解这个三角形。

解:在$\mathrm{Rt}△ABC$中,$∠ C=90°,a=4\sqrt{15},c=8\sqrt{5}$,
$\therefore \sin A = \frac{a}{c} = \frac{4\sqrt{15}}{8\sqrt{5}} = \frac{\sqrt{3}}{2},\therefore ∠ A = 60°$,
$\therefore ∠ B = 90° - 60° = 30°,\therefore b = \frac{1}{2}c = 4\sqrt{5}$.
故$b=4\sqrt{5},∠ A=60°,∠ B=30°$.

3. [2026 滁州天长实验中学月考]已知在$Rt△ ABC$中，$∠ C=90°$，$∠ A=α$，$AC=m$，那么$AB$的长为（）

A.$m\sinα$
B.$m\cosα$
C.$\dfrac{m}{\cosα}$
D.$\dfrac{m}{\sinα}$

在$\mathrm{Rt}△ABC$中,$∠ C=90°,\therefore \cos A = \frac{AC}{AB}$,即$\cos α = \frac{m}{AB},\therefore AB = \frac{m}{\cos α}$.

4. [2025 芜湖期末] 在$△ ABC$中，$∠ ABC = 90°$，$∠ BAC = 60°$，$BC = 4$，则$AB$的长为 （）

A.$4$
B.$8$
C.$4\sqrt{3}$
D.$\dfrac{4}{3}\sqrt{3}$

在$△ ABC$中,$∠ ABC=90°,∠ BAC=60°,\therefore \tan ∠ BAC = \frac{BC}{AB} = \sqrt{3},\therefore AB = \frac{BC}{\sqrt{3}} = \frac{4}{3}\sqrt{3}$.

变式 易错题 [2024 哈尔滨中考] △ABC是直角三角形，AB = 2√3，∠ABC = 30°，则AC的长为。

分情况讨论如下:若$∠ A = 90°$,则$\tan ∠ ABC = \frac{AC}{AB},\therefore AC = AB · \tan ∠ ABC = 2\sqrt{3} × \frac{\sqrt{3}}{3} = 2$;若$∠ C = 90°$,则$\sin ∠ ABC = \frac{AC}{AB},\therefore AC = AB · \sin ∠ ABC = 2\sqrt{3} × \frac{1}{2} = \sqrt{3}$.

5. [2025西藏中考]把一副直角三角板按图1的位置摆放，抽象成如图2的示意图，已知DC = 6 cm，求四边形ABCD的面积(结果保留根号)。

解:由题意可知,$DC=6\ \mathrm{cm},∠ DBC=30°,∠ BDC=90°$,
$\therefore \tan ∠ DBC = \frac{DC}{BD} = \frac{\sqrt{3}}{3},\therefore BD = \sqrt{3}DC = 6\sqrt{3}$.
在$\mathrm{Rt}△ABD$中,$∠ BAD=90°,∠ DBA=45°,AB=AD$,
$\therefore \sin ∠ DBA = \frac{AD}{BD} = \frac{\sqrt{2}}{2},\therefore AB = AD = \frac{\sqrt{2}}{2}BD = 3\sqrt{6}$,
$\therefore S_{\mathrm{四边形}ABCD} = S_{△ ABD} + S_{△ BDC} = \frac{1}{2}AB · AD + \frac{1}{2}BD · DC = \frac{1}{2} × 3\sqrt{6} × 3\sqrt{6} + \frac{1}{2} × 6\sqrt{3} × 6 = (27 + 18\sqrt{3})(\mathrm{cm}^2)$.

6. [2026 合肥蜀山区段考] 如图，在Rt△ABC中，∠ACB = 90°，D是AB的中点，CD = $\frac{5}{2}$，AC = 4，则tan∠DCB的值是（）

A.$\frac{4}{5}$
B.$\frac{3}{5}$
C.$\frac{4}{3}$
D.$\frac{3}{4}$

$\because ∠ ACB=90°$,$CD$是$AB$边上的中线,$\therefore BD=CD$,$AB=2CD=5$,$\therefore ∠ B = ∠ DCB$. 在$\mathrm{Rt}△ABC$中,由勾股定理,得$BC = \sqrt{AB^2 - AC^2} = \sqrt{5^2 - 4^2} = 3$,$\therefore \tan ∠ DCB = \tan ∠ B = \frac{AC}{BC} = \frac{4}{3}$.

7. [2025 乐山中考] 如图, 在 $△ ABC$ 中, $∠ ABC = 45°, ∠ ACB = 60°, AC = 2$.
(1)求 AB 的长;
(2)求点 C 到线段 AB 的距离.

解:(1)如图,过点$A$作$AJ ⊥ BC$于点$J$.
在$\mathrm{Rt}△ACJ$中,$AC=2$,$∠ ACJ=60°$,
$\therefore AJ = AC · \sin 60° = \sqrt{3}$,$CJ = AC · \cos 60° = 1$.
在$\mathrm{Rt}△ABJ$中,$∠ B=45°$,$\therefore ∠ BAJ=45°$,
$\therefore BJ = AJ = \sqrt{3}$,
$\therefore AB = \frac{AJ}{\sin B} = \sqrt{2}AJ = \sqrt{6}$.

(2)如图,过点$C$作$CK ⊥ AB$于点$K$.
由(1)可知$BC = BJ + CJ = 1 + \sqrt{3}$,
$\therefore CK = BC · \sin B = (1 + \sqrt{3}) × \frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{2}$,
$\therefore$ 点$C$到线段$AB$的距离为$\frac{\sqrt{2} + \sqrt{6}}{2}$.