12. [2026宿州段考] 如图,在菱形ABCD中,DE⊥AB,sin A = $\frac{4}{5}$, BE=2,则cos∠DBE = ()

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

12 B 在$△ADE$中,$∠AED = 90°$,$\sin A = \frac{4}{5}$,$\therefore \sin A = \frac{DE}{AD} = \frac{4}{5}$,$\therefore$ 设$DE = 4x, AD = 5x$,由勾股定理,得$AE = \sqrt{AD^2 - DE^2} = 3x$. $\because BE = 2$,$\therefore AB = 3x + 2$. 由菱形的性质,可知$AB = AD$,$\therefore 3x + 2 = 5x$,解得$x = 1$,$\therefore DE = 4$. 在$\mathrm{Rt}△BDE$中,由勾股定理,得$BD = \sqrt{DE^2 + BE^2} = \sqrt{4^2 + 2^2} = 2\sqrt{5}$,$\therefore \cos∠DBE = \frac{BE}{BD} = \frac{2}{2\sqrt{5}} = \frac{\sqrt{5}}{5}$.

13. [2026合肥包河区期中]如图,在网格中,小正方形的边长均为1,点A,B,C都在格点上,连接AB,BC,则∠ABC的余弦值是 ()

A.2
B.$\frac{2\sqrt{5}}{5}$
C.$\frac{\sqrt{5}}{5}$
D.$\frac{1}{2}$

13 B 解题思路：

14. [2025合肥包河区段考]如图,在$△ ABC$中,$CD ⊥ AB, BE ⊥ AC$,若$\frac{DE}{BC}=\frac{2}{5}$,则$\sin A$的值为。

14 $\frac{\sqrt{21}}{5}$ $\because CD ⊥ AB, BE ⊥ AC$,$\therefore ∠AEB = ∠ADC$.
又$\because ∠A = ∠A$,$\therefore △ABE ∽ △ACD$,$\therefore \frac{AD}{AE} = \frac{AC}{AB}$,$\therefore \frac{AD}{AC} = \frac{AE}{AB}$.
又$\because ∠A = ∠A$,$\therefore △AED ∽ △ABC$,$\therefore \frac{AD}{AC} = \frac{DE}{BC} = \frac{2}{5}$. 设$AD = 2a, AC = 5a$,在$\mathrm{Rt}△ACD$中,由勾股定理得$CD = \sqrt{AC^2 - AD^2} = \sqrt{21}a$,$\therefore \sin A = \frac{CD}{AC} = \frac{\sqrt{21}}{5}$.

15. 如图,在$△ ABC$中,$∠ C=90°$,点$D,E$分别在$AC,AB$上,$BD$平分$∠ ABC$,$DE ⊥ AB$于点$E$,$AE=6$,$\cos A=\dfrac{3}{5}$.
(1)求$CD$的长;
(2)求$\tan∠ DBC$的值.

15 解:(1)在$\mathrm{Rt}△ADE$中,$AE = 6$,$\cos A = \frac{3}{5}$,
所以$AD = \frac{AE}{\cos A} = 10$,
由勾股定理,得$DE = \sqrt{AD^2 - AE^2} = \sqrt{10^2 - 6^2} = 8$.
因为$DE ⊥ AB$,$DC ⊥ BC$,$BD$平分$∠ABC$,
所以由角平分线的性质,得$CD = DE = 8$.
(2)由(1)得$AC = 18$,
在$\mathrm{Rt}△ACB$中,$\cos A = \frac{AC}{AB} = \frac{3}{5}$,所以$AB = 30$,
由勾股定理,得$BC = \sqrt{AB^2 - AC^2} = \sqrt{30^2 - 18^2} = 24$,
所以$\tan∠DBC = \frac{CD}{BC} = \frac{8}{24} = \frac{1}{3}$.

16. 运算能力 [2025 淄博调研]如果三角形有一边上的中线长等于这边的长,那么称这个三角形为“好玩三角形”.若$\mathrm{Rt}△ ABC$是“好玩三角形”且$∠ A=90°$,则$\cos∠ ABC$的值为.

16 $\frac{2\sqrt{7}}{7}$或$\frac{\sqrt{21}}{7}$ 根据直角三角形斜边上的中线等于斜边的一半,可知当一边上的中线长等于这边的长时,这条边一定是直角边,故分两种情况讨论:①如图1,在$\mathrm{Rt}△ABC$中,$∠A = 90°$,$CE$是$△ABC$的中线,设$AB = EC = 2a$,则$AE = EB = a$,$\therefore AC = \sqrt{3}a$,$\therefore BC = \sqrt{7}a$,$\therefore \cos∠ABC = \frac{AB}{BC} = \frac{2a}{\sqrt{7}a} = \frac{2\sqrt{7}}{7}$.
②如图2,在$\mathrm{Rt}△ABC$中,$∠A = 90°$,$BE$是$△ABC$的中线,设$EB = AC = 2b$,则$AE = EC = b$,$\therefore AB = \sqrt{3}b$,$\therefore BC = \sqrt{7}b$,$\therefore \cos∠ABC = \frac{AB}{BC} = \frac{\sqrt{3}b}{\sqrt{7}b} = \frac{\sqrt{21}}{7}$.综上,$\cos∠ABC$的值为$\frac{2\sqrt{7}}{7}$或$\frac{\sqrt{21}}{7}$.

1. 一题多解 [2026宿州十一中质检] 如图，在平面直角坐标系$xOy$中，有三点$A(0,1),B(4,1),C(5,6)$，则$\sin∠ BAC=$ ()

A.$\frac{1}{2}$
B.$\frac{\sqrt{13}}{5}$
C.$\frac{\sqrt{2}}{2}$
D.$\frac{\sqrt{3}}{2}$

1 C 通解 如图1,过点$C$作$CD ⊥ AB$交$AB$的延长线于点$D$,$\because A(0,1),B(4,1),C(5,6)$,$\therefore D(5,1)$,$\therefore CD = 5$,$AD = 5$,由勾股定理易得$AC = 5\sqrt{2}$,$\therefore \sin∠BAC = \frac{CD}{AC} = \frac{\sqrt{2}}{2}$.
另解一 如图2,过点$B$作$DB ⊥ AB$交$AC$于点$D$,$\because A(0,1),B(4,1),C(5,6)$,$\therefore D(4,5)$,$\therefore BD = 4$,$AB = 4$,由勾股定理易得$AD = 4\sqrt{2}$,$\therefore \sin∠BAC = \sin∠BAD = \frac{BD}{AD} = \frac{\sqrt{2}}{2}$.
另解二 如图3,在$AC$上取格点$D(2,3)$,连接$BD$,结合$A(0,1),B(4,1)$,易得$AD = 2\sqrt{2}$,$BD = 2\sqrt{2}$,$AB = 4$,$\therefore AD^2 + BD^2 = AB^2$,$\therefore ∠ADB = 90°$(点拨:勾股定理的逆定理),$\therefore \sin∠BAC = \sin∠BAD = \frac{BD}{AB} = \frac{\sqrt{2}}{2}$.

2. [2026蚌埠期末]如图,在由正三角形构成的网格图中,A,B,C三点均在格点上,则$\cos∠ BAC$的值为.

2 $\frac{2\sqrt{7}}{7}$ 解题思路:根据等边三角形的性质可得$∠ACB = 90°$,设正三角形的边长为1,分别求出$BC$,$AB$的长,最后根据三角函数定义即可求出$\cos∠BAC$.
如图,由正三角形的性质可知$∠ACB = 60° + \frac{1}{2} × 60° = 90°$.
设正三角形的边长为1,在$\mathrm{Rt}△BCE$中,$BE = 2$,$EC = 1$,由勾股定理,得$BC = \sqrt{BE^2 - EC^2} = \sqrt{2^2 - 1^2} = \sqrt{3}$. 在$\mathrm{Rt}△ABC$中,$AC = 2$,$BC = \sqrt{3}$,由勾股定理,可得$AB = \sqrt{AC^2 + BC^2} = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{7}$,$\therefore \cos∠BAC = \frac{AC}{AB} = \frac{2}{\sqrt{7}} = \frac{2\sqrt{7}}{7}$.

3. [2026安庆四中月考]如图,在正方形方格纸中,每个小正方形的边长都相等,A,B,C,D都在格点处,AB与CD相交于点P,则$\cos∠ APC$的值为。

3 $\frac{2\sqrt{5}}{5}$ 把$AB$向上平移一个单位长度到$DE$,连接$CE$,如图,则$DE // AB$,$\therefore ∠APC = ∠EDC$. 设小正方形的边长为1,则$EC = \sqrt{2^2 + 1^2} = \sqrt{5}$,$DC = \sqrt{4^2 + 2^2} = 2\sqrt{5}$,$DE = \sqrt{3^2 + 4^2} = 5$.
$\because EC^2 + DC^2 = DE^2$,$\therefore △DCE$为直角三角形,$∠DCE = 90°$,
$\therefore \cos∠APC = \cos∠EDC = \frac{DC}{DE} = \frac{2\sqrt{5}}{5}$(等角转化法).

提分大招
当网格中所求角不在直角三角形中,常采用以下方法构造直角三角形解决.
①作垂线法:作垂线直接构造顶点在格点上的直角三角形.
②等角转化法:当网格中没有特殊的直线,或不易构造以格点为顶点的直角三角形时,常常通过构造平行线或余角转化的方式转化到易求三角函数值的等角中解决.