6. [2025安庆段考]在正方形网格中，∠AOB如图放置，则$\sin∠ AOB$的值为（）


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

6 B 解题思路:求锐角的三角函数值,需构造直角三角形,根据网格特点,由勾股定理求出线段长,进而根据等腰三角形“三线合一”的性质,得到直角三角形,再根据三角函数的定义求解.
如图,取格点 $C,D$,连接$AD,CD$,设小正方形的边长均是1,根据勾股定理可得 $OD = AD = \sqrt{10}$,$OC = AC = \sqrt{5}$,在 $△ ODA$ 中,由等腰三角形三线合一,得 $∠ OCD = 90°$, 则 $CD = \sqrt{OD^2 - OC^2} = \sqrt{5}$,$\therefore \sin ∠ AOB = \frac{CD}{OD} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{\sqrt{2}}{2}$.

7. [2025马鞍山期末]如图,将以A为直角顶点的等腰直角三角形ABC沿直线BC平移得到△A'B'C',使点B'与点C重合,连接A'B,则sin∠A'BC'的值为。

7 $\frac{\sqrt{10}}{10}$ 如图,过点 $A'$ 作出 $A'D ⊥ BC'$,垂足为 $D$. 由题意可知,$△ A'B'C'$ 是等腰直角三角形,$\therefore B'D = \frac{1}{2}B'C'$,$∠ B'A'D = \frac{1}{2}∠ B'A'C' = 45°$,易知 $△ A'B'D$ 也是等腰直角三角形,$\therefore A'D = B'D = \frac{1}{2}B'C'$. 由平移的性质可知,$B'C' = BC$,$\therefore BD = BC + B'D = B'C' + \frac{1}{2}B'C' = \frac{3}{2}B'C'$,$\therefore A'B = \sqrt{A'D^2 + BD^2} = \frac{\sqrt{10}}{2}B'C'$,$\therefore \sin ∠ A'BC' = \frac{A'D}{A'B} = \frac{\frac{1}{2}B'C'}{\frac{\sqrt{10}}{2}B'C'} = \frac{\sqrt{10}}{10}$.

8. 一题多解 如图,在$Rt△ ABC$中,$∠ C=90°$,D为BC上一点,$AB=5$,$BD=1$,$\tan B=\frac{3}{4}$.
(1)求AD的长;
(2)求$\sin α$的值.

8 解:(1)在 $\mathrm{Rt}△ABC$ 中,$∠ C=90°$,$\tan B = \frac{AC}{BC} = \frac{3}{4}$,设 $AC = 3x(x ＞ 0)$,则 $BC = 4x$,由勾股定理,得 $AC^2 + BC^2 = AB^2$,即 $(3x)^2 + (4x)^2 = 5^2$,$\therefore x = 1$,$\therefore AC = 3$,$BC = 4$.
$\because BD = 1$,$\therefore CD = 3$,
$\therefore AD = \sqrt{CD^2 + AC^2} = \sqrt{3^2 + 3^2} = 3\sqrt{2}$.
(2) 通解 如图,过点 $D$ 作 $DE ⊥ AB$ 于点 $E$,
$\sin B = \frac{AC}{AB} = \frac{DE}{BD}$,即 $\frac{3}{5} = \frac{DE}{1}$,
$\therefore DE = \frac{3}{5}$,
$\therefore \sin α = \frac{DE}{AD} = \frac{\frac{3}{5}}{3\sqrt{2}} = \frac{\sqrt{2}}{10}$.
另解 如图,过点 $B$ 作 $BM ⊥ AD$,交 $AD$ 的延长线于点 $M$.
由(1)得,$CD = AC = 3$,$\therefore ∠ ADC = 45°$,$\therefore ∠ BDM = 45°$.
在 $\mathrm{Rt}△BMD$ 中,$BD = 1$,
$\therefore BM = BD · \sin ∠ BDM = 1 × \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$.
$\therefore$ 在 $\mathrm{Rt}△BMA$ 中,$\sin α = \frac{BM}{AB} = \frac{\frac{\sqrt{2}}{2}}{5} = \frac{\sqrt{2}}{10}$.

9. [2025黑龙江龙东地区一模]如图,四边形ABCD为菱形,对角线AC,BD交于点O,DE⊥AB,垂足为E.若AB=10,BD=12,则$\cos∠ EDB$为 ()


A.$\frac{4}{5}$
B.$\frac{3}{5}$
C.$\frac{3}{4}$
D.$\frac{1}{2}$

9 A $\because$ 四边形 $ABCD$ 为菱形,$\therefore OB = \frac{1}{2}BD = 6$,$AC ⊥ BD$,$\therefore ∠ AOB = ∠ OAB + ∠ OBA = 90°$,$\therefore OA = \sqrt{AB^2 - OB^2} = \sqrt{10^2 - 6^2} = 8$.$\because DE ⊥ AB$,$\therefore ∠ EDB + ∠ DBE = 90°$.$\because ∠ DBE = ∠ OBA$,$\therefore ∠ EDB = ∠ OAB$,$\therefore \cos ∠ EDB = \cos ∠ OAB = \frac{OA}{AB} = \frac{8}{10} = \frac{4}{5}$.

10. [2026亳州段考]如图,在边长为1的小正方形组成的网格中,点A,B,C,D,E都在网格的格点上,则∠ADC的正弦值为 ()


A.$\frac{\sqrt{10}}{2}$
B.$\frac{1}{3}$
C.$\frac{2}{3}$
D.$\frac{\sqrt{10}}{10}$

10 D 解题思路:先证明$△ ACD ≌ △ BCE$,从而可得$∠ ADC = ∠ BEC$,然后求出$\sin ∠ BEC$的值,即可得解.
如图,易得 $AC = BC = \sqrt{5}$,$CD = CE = \sqrt{10}$,$AD = BE = 5$,$\therefore △ ACD ≌ △ BCE(\mathrm{SSS})$,$\therefore ∠ ADC = ∠ BEC$. 在 $\mathrm{Rt}△EFC$ 中,$FC = 1$,$EC = \sqrt{10}$,$\therefore \sin ∠ BEC = \frac{FC}{CE} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$,$\therefore \sin ∠ ADC = \frac{\sqrt{10}}{10}$.

11. 如图,在$△ ABC$中,$AB = AC = 5$,$BC = 8$.若$∠ BPC = \frac{1}{2}∠ BAC$,则$\tan∠ BPC =$ .

11 $\frac{4}{3}$ 如图,过点 $A$ 作 $AE ⊥ BC$ 于点 $E$. 因为 $AB = AC$,所以$BE = \frac{1}{2}BC = 4$,$∠ BAE = \frac{1}{2}∠ BAC$,故 $∠ BPC = ∠ BAE$. 在$\mathrm{Rt}△BAE$中,由勾股定理,得 $AE = \sqrt{AB^2 - BE^2} = \sqrt{5^2 - 4^2} = 3$,所以 $\tan ∠ BPC = \tan ∠ BAE = \frac{BE}{AE} = \frac{4}{3}$.

12. 如图,已知E是矩形ABCD的对角线AC上的一动点,正方形EFGH的顶点G,H都在边AD上,AB = 3, BC = 4, 则 $\tan ∠ AFE$ 的值是.

12 $\frac{3}{7}$ $\because EH // CD$,$\therefore △ AEH ∽ △ ACD$,$\therefore \frac{EH}{AH} = \frac{CD}{AD} = \frac{3}{4}$. 设$EH = 3a(a ＞ 0)$,则 $AH = 4a$,$\therefore HG = GF = 3a$,$\because EF // AD$,$\therefore ∠ AFE = ∠ FAG$,$\therefore \tan ∠ AFE = \tan ∠ FAG = \frac{GF}{AG} = \frac{3a}{3a + 4a} = \frac{3}{7}$.