1. 在$△ ABC$中，$∠ A$，$∠ B$，$∠ C$的对边分别为$a$，$b$，$c$，且$a:b:c=3:4:5$. 求证：$\sin A + \sin B = \frac{7}{5}$.

证明:由题意设 $a=3k,b=4k,c=5k$,其中 $k＞0$,
因为 $(3k)^2 + (4k)^2 = (5k)^2$,即 $a^2 + b^2 = c^2$,
所以 $△ ABC$ 是直角三角形,且 $∠ C = 90°$.
则 $\sin A = \frac{a}{c} = \frac{3k}{5k} = \frac{3}{5}$ ,$\sin B = \frac{b}{c} = \frac{4k}{5k} = \frac{4}{5}$.
所以 $\sin A + \sin B = \frac{7}{5}$.
避坑指南
本题的易错之处是没有说明 $∠ C = 90°$,而直接利用正弦的定义进行证明. 表面上看,结果似乎没有错,但推理过程中,却缺少了依据. 解答此类题时应先判断三角形的形状,再利用正弦、余弦的定义进行解答.

2. 在$\mathrm{Rt}△ ABC$中，$AC=8$，$BC=6$，则$\cos A$的值为（）

A.$\frac{3}{5}$
B.$\frac{\sqrt{7}}{4}$
C.$\frac{4}{5}$或$\frac{\sqrt{7}}{4}$
D.$\frac{4}{5}$或$\frac{2\sqrt{7}}{7}$

①当$∠ C = 90°$时,由勾股定理,得$AB = \sqrt{AC^2 + BC^2} = 10$,$\therefore \cos A = \frac{AC}{AB} = \frac{8}{10} = \frac{4}{5}$;②当$∠ B = 90°$时,由勾股定理,得$AB = \sqrt{AC^2 - BC^2} = 2\sqrt{7}$,$\therefore \cos A = \frac{AB}{AC} = \frac{2\sqrt{7}}{8} = \frac{\sqrt{7}}{4}$. 综上所述,$\cos A$的值为$\frac{4}{5}$或$\frac{\sqrt{7}}{4}$.

3. [2025哈工大附中月考]已知正方形ABCD的边长为2，P是直线CD上一点，若DP = 1，则$\tan∠ BPC$的值是.

①如图1, $\because BC = CD = 2$,$DP = 1$,$∠ C = 90°$,
$\therefore PC = 1$,$\therefore \tan∠ BPC = \frac{BC}{PC} = 2$. ②如图2, $\because DP = 1$,$CD = 2$,
$\therefore PC = 3$. 又$\because BC = 2$,$∠ C = 90°$,$\therefore \tan∠ BPC = \frac{BC}{PC} = \frac{2}{3}$. 综上所述,$\tan∠ BPC$的值是2或$\frac{2}{3}$.
避坑指南
对于条件“P是直线CD上一点”中的“直线”理解不清,往往会只考虑“点P在线段CD上”这一种情况,而忽略“点P在线段CD的延长线上”这一种情况.

4. [2025池州期末]在$△ ABC$中，已知AD是$△ ABC$的高，$CD = 1$，$AD = BD = \sqrt{3}$，则$∠ BAC$的度数为。

①当垂足D在线段BC上时,如图1所示. 在$\mathrm{Rt}△ ABD$中,$\tan∠ BAD = \frac{BD}{AD} = 1$,$\therefore ∠ BAD = 45°$. 在$\mathrm{Rt}△ ACD$中,$\tan∠ CAD = \frac{CD}{AD} = \frac{\sqrt{3}}{3}$,$\therefore ∠ CAD = 30°$. $\therefore ∠ BAC = ∠ BAD + ∠ CAD = 45° + 30° = 75°$.
②当垂足D在线段BC的延长线上时,如图2所示. 在$\mathrm{Rt}△ ABD$中,$\tan∠ BAD = \frac{BD}{AD} = 1$,$\therefore ∠ BAD = 45°$. 在$\mathrm{Rt}△ CAD$中,$\tan∠ CAD = \frac{CD}{AD} = \frac{\sqrt{3}}{3}$,$\therefore ∠ CAD = 30°$. $\therefore ∠ BAC = ∠ BAD - ∠ CAD = 45° - 30° = 15°$. 综上,$∠ BAC$的度数为$75°$或$15°$.
避坑指南
本题的易错之处是忽略高在$△ ABC$外部的情况. 当涉及三角形的高(或面积),而没有给出图形时,常要考虑两种情况:高在三角形内或高在三角形外.

5. 在$△ ABC$中，$AC=2\sqrt{5}$，点$D$为直线$AB$上一点，且$AB=3BD$，直线$CD$与直线$BC$所夹锐角的正切值为$\frac{1}{3}$，且$CD ⊥ AC$，求$BC$的长。

解:①如图1,过点$B$作$BM⊥ CD$,垂足为$M$.
$\because CD⊥ AC$,$\therefore BM// AC$,
$\therefore △ DBM∽△ DAC$,$\therefore \frac{BM}{AC} = \frac{DB}{DA}$.
$\because AB = 3BD$,$AC = 2\sqrt{5}$,
$\therefore \frac{BM}{2\sqrt{5}} = \frac{1}{4}$,$\therefore BM = \frac{\sqrt{5}}{2}$.
在$\mathrm{Rt}△ BMC$中,由$\tan∠ BCM = \frac{1}{3}$,
可得$BC = \sqrt{10}BM = \frac{5\sqrt{2}}{2}$.
②如图2,过点$B$作$BN⊥ AC$,交$AC$的延长线于点$N$.
$\because CD⊥ AC$,
$\therefore BN// CD$,$\therefore \frac{AC}{CN} = \frac{AD}{DB}$.
$\because AB = 3BD$,$AC = 2\sqrt{5}$,
$\therefore \frac{2\sqrt{5}}{CN} = \frac{2}{1}$,$\therefore CN = \sqrt{5}$.
由$BN// CD$,得$∠ CBN = ∠ BCD$.
$\because \tan∠ BCD = \frac{1}{3}$,$\therefore \tan∠ CBN = \frac{1}{3}$,$\therefore BC = \sqrt{10}CN = 5\sqrt{2}$.
综上,$BC$的长为$\frac{5\sqrt{2}}{2}$或$5\sqrt{2}$.

6. 跨学科·物理 [2025泰州二模]如图，CD是平面镜，光线从A点出发经CD上点O反射后照射到B点，若入射角为α，反射角为β(反射角等于入射角)，AC⊥CD于点C，BD⊥CD于点D，且AC=3，BD=6，CD=12，则$\tan α$的值为 ()

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

解题思路:先根据平行线的性质得到$∠ A = α$,$∠ B = β$,从而可得$∠ A = ∠ B$,再判定$△ AOC∽△ BOD$,根据相似三角形的性质可得$OC$的长,根据正切的定义即可求得$\tan α$.
如图,由题意,得$OE⊥ CD$. $\because AC⊥ CD$,$\therefore AC// OE$,$\therefore ∠ A = α$,同理可得$∠ B = β$. $\because α = β$(反射角等于入射角),$\therefore ∠ A = ∠ B$. 又$\because ∠ ACO = ∠ BDO$,$\therefore △ AOC∽△ BOD$,$\therefore \frac{OC}{OD} = \frac{AC}{BD}$,$\therefore \frac{OC}{12-OC} = \frac{3}{6}$,解得$OC=4$,$\therefore \tan α = \tan A = \frac{OC}{AC} = \frac{4}{3}$.

7. [2025合肥调研]如图,四边形ABCD的对角线AC平分∠BAD,ED⊥AD,BC⊥AC,且$\cos∠ CBE = \dfrac{15}{16}$,$∠ ABE = 30°$,则$\dfrac{AD}{AC}$的值为（）

A.$\dfrac{\sqrt{3}}{3}$
B.$\dfrac{\sqrt{3} - 1}{2}$
C.$\dfrac{1}{2}$
D.$\dfrac{8}{15}$

解题思路:过点$E$作$EF⊥ AB$,先根据角平分线的定义和性质可得$∠ DAE = ∠ CAB$,$EF = ED$,再根据含$30°$角的直角三角形的性质可得$BE = 2EF$,再由锐角三角函数的定义可得$\frac{BC}{BE} = \frac{15}{16}$,整理得$\frac{DE}{BC} = \frac{8}{15}$,最后证明$△ ADE∽△ ACB$,然后利用相似三角形的性质列比例式即可求解.
如图,过点$E$作$EF⊥ AB$,垂足为$F$.
$\because AC$平分$∠ BAD$,$ED⊥ AD$,$\therefore ∠ DAE = ∠ CAB$,$EF = ED$. $\because ∠ EFB = 90°$,$∠ ABE = 30°$,$\therefore BE = 2EF$. 在$\mathrm{Rt}△ BCE$中,$\cos∠ CBE = \frac{BC}{BE} = \frac{15}{16}$,$\therefore \frac{BC}{DE} = \frac{15}{8}$,即$\frac{DE}{BC} = \frac{8}{15}$. $\because ∠ EAD = ∠ BAC$,$∠ ADE = ∠ ACB$,$\therefore △ ADE∽△ ACB$,$\therefore \frac{AD}{AC} = \frac{DE}{BC} = \frac{8}{15}$.

8. [2025威海中考]【问题提出】已知$∠ α$，$∠ β$都是锐角，$\tan α =\dfrac{1}{2}$，$\tan β =\dfrac{1}{3}$，求$∠ α + ∠ β$的度数.
【问题解决】(1)如图，小亮同学在边长为1的正方形网格中画出$∠ BAD$和$∠ CAD$，请你按照这个思路求$∠ α + ∠ β$的度数为________.(点$A,B,C,D$都在格点上)

【策略迁移】(2)已知$∠ α$，$∠ β$都是锐角，$\tan α =\dfrac{2}{3}$，$\tan β =\dfrac{3}{2}$，则$∠ α + ∠ β = \_\_\_\_\_\_°$.
(3)已知$∠ α$，$∠ β$，$∠ θ$都是锐角，$\tan α =\dfrac{1}{3}$，$\tan β =\dfrac{1}{7}$，$∠ α + ∠ β = ∠ θ$，则$\tan θ$的值为________.
(提示:在正方形网格中画出求解过程的图形)

(1)$45°$
如图1,连接$BC$,$\because AB = BC = \sqrt{5}$,$AC = \sqrt{10}$,$\therefore AB^2 + BC^2 = AC^2$,$\therefore ∠ ABC = 90°$(依据:勾股定理的逆定理),$\therefore ∠ BAC = 45°$,$\therefore ∠ α + ∠ β = 45°$.
(2)$90$
如图2,连接$BC$,由题意,得$∠ α = ∠ BAD$,$∠ β = ∠ DAC$,$\because △ ABC$是等腰直角三角形,$\therefore ∠ α + ∠ β = 90°$.
(3)$\frac{1}{2}$
如图2,$∠ α = ∠ GDH$,$∠ β = ∠ HDF$,$FG = \sqrt{10}$,$DG = 2\sqrt{10}$,$DF = 5\sqrt{10}$,$\because FG^2 + DG^2 = DF^2$,$\therefore △ DFG$为直角三角形.
在$\mathrm{Rt}△ DGF$中,$\tan θ = \tan(α+β) = \frac{FG}{DG} = \frac{1}{2}$.