9. (2025·淄博期中)如图是由边长相同的小正方形组成的网格，A,B,P,Q四点均在正方形网格的格点上，线段AB,PQ相交于点M，则图中$\cos∠ QMB$的值是 ()

A.$\dfrac{\sqrt{5}}{5}$
B.$\dfrac{2\sqrt{5}}{5}$
C.$\dfrac{\sqrt{2}}{2}$
D.$\dfrac{\sqrt{10}}{5}$

9. A 解析:如图所示,平移AB使A点与P点重合,连接B'Q,可得∠QMB=∠P.
∵ PB'= $\sqrt{2^2+2^2}$=2$\sqrt{2}$,PQ= $\sqrt{2^2+6^2}$=2$\sqrt{10}$,B'Q= $\sqrt{4^2+4^2}$=4$\sqrt{2}$,
∴ PB'$^2$+QB'$^2$=PQ$^2$,
∴ △QPB'是直角三角形,
∴ cos∠QMB=cos P = $\frac{B'P}{PQ}$ = $\frac{2\sqrt{2}}{2\sqrt{10}}$=$\frac{\sqrt{5}}{5}$.故选 A.

10. (2026·珠海模拟)如图，AD是$\mathrm{Rt}△ ABC$中直角的平分线，$BC = 5$，$\cos B = 0.8$，则$\frac{S_{△ ACD}}{S_{△ ABD}}=$

10. $\frac{3}{4}$ 解析:如图,在Rt△ABC中,∠BAC=90°,BC=5,cos B=0.8,
∴ AB=BC·cos B=5×0.8=4,由勾股定理得AC= $\sqrt{BC^2-AB^2}$ = $\sqrt{5^2-4^2}$=3,过点D作DE⊥AB于点E,DF⊥AC于点F.
∵ AD平分∠BAC,DE⊥AB,DF⊥AC,
∴ DE=DF,
∴ $\frac{S_{△ ACD}}{S_{△ ABD}}$ = $\frac{\frac{1}{2}AC·DF}{\frac{1}{2}AB·DE}$ = $\frac{AC}{AB}$ = $\frac{3}{4}$.

11. (扬州中考)在$△ ABC$中,$∠ C=90°$,$a,b,c$分别为$∠ A,∠ B,∠ C$的对边,若$b^2=ac$,则$\sin A$的值为.

11. $\frac{-1+\sqrt{5}}{2}$ 解析:在Rt△ABC中,由勾股定理可知 a$^2$ + b$^2$ = c$^2$.
∵ ac = b$^2$,
∴ a$^2$ + ac = c$^2$.
∵ a＞0,b＞0,c＞0,
∴ $\frac{a^2+ac}{c^2}$ = $\frac{c^2}{c^2}$,即 $(\frac{a}{c})^2$ + $\frac{a}{c}$ = 1,求出 $\frac{a}{c}$ = $\frac{-1+\sqrt{5}}{2}$ 或 $\frac{a}{c}$ = $\frac{-1-\sqrt{5}}{2}$ (舍去),
∴ 在Rt△ABC中,sin A = $\frac{a}{c}$ = $\frac{-1+\sqrt{5}}{2}$.

12. (2025·兰州中考)如图，$\odot O$是$△ ABC$的外接圆，$AB$是$\odot O$的直径，过点$B$的切线交$AC$的延长线于点$D$，连接$DO$并延长，交$\odot O$于点$E$，连接$AE$，$CE$.
(1)求证：$∠ ADB = ∠ AEC$；
(2)若$AB=4$，$\cos∠ AEC=\dfrac{\sqrt{5}}{3}$，求$OD$的长.

12. (1)
∵ BD是⊙O的切线,
∴ ∠ABD=90°.
∵ AB是⊙O的直径,
∴ ∠ACB=90°,
∴ ∠ADB=90°-∠DBC=∠ABC.
∵ $\overset{\frown}{AC}$=$\overset{\frown}{AC}$,
∴ ∠AEC=∠ABC,
∴ ∠ADB=∠AEC.
(2) 由(1)得∠ADB=∠ABC=∠AEC,
∴ cos∠ADB=cos∠ABC=cos∠AEC,
∴ $\frac{BD}{AD}$ = $\frac{BC}{AB}$ = $\frac{\sqrt{5}}{3}$.
∵ AB=4,
∴ BC=$\frac{4\sqrt{5}}{3}$,
∴ AC= $\sqrt{AB^2-BC^2}$ = $\sqrt{4^2-(\frac{4\sqrt{5}}{3})^2}$ = $\frac{8}{3}$.
∵ ∠ADB=∠ABC,∠ABD=∠ACB=90°,
∴ △ADB∽△ABC,
∴ $\frac{AB}{AD}$ = $\frac{AC}{AB}$,即 $\frac{4}{AD}$ = $\frac{\frac{8}{3}}{4}$,解得 AD=6,
∴ BD$^2$=AD$^2$-AB$^2$=36-16=20.
∵ OB=$\frac{1}{2}$AB=2,
∴ OD= $\sqrt{OB^2+BD^2}$ = $\sqrt{2^2+20}$=2$\sqrt{6}$.

13. 如图,在$△ ABC$中,$AB = AC$,$∠ A = 36°$,$BD$平分$∠ ABC$,交$AC$于点$D$,则$\cos A = (\quad)$


A.$\dfrac{\sqrt{5}-1}{4}$
B.$\dfrac{\sqrt{5}+1}{4}$
C.$\dfrac{\sqrt{5}-1}{2}$
D.$\dfrac{3-\sqrt{5}}{2}$

13. B 解析:如图所示,过点D作DE⊥AB于点E,设AB=AC=a,BC=b.
∵ AB=AC,∠A=36°,
∴ ∠ABC=∠C=$\frac{180°-∠A}{2}$=72°.
∵ BD平分∠ABC,
∴ ∠ABD=∠CBD=$\frac{1}{2}$∠ABC=36°,
∴ ∠A=∠CBD=∠ABD,∠BDC=∠A+∠ABD=72°,
∴ ∠BDC=∠C,AD=BD=BC=b,
∴ DC=AC-AD=a-b.
∵ DE⊥AB,
∴ AE=$\frac{1}{2}$AB=$\frac{1}{2}$a.
∵ ∠ACB=∠BCD,
∴ △ABC∽△BDC,
∴ $\frac{AB}{BD}$ = $\frac{BC}{DC}$,即 $\frac{a}{b}$ = $\frac{b}{a-b}$,
∴ 用a表示b得b$_1$=$\frac{\sqrt{5}-1}{2}$a,b$_2$=$\frac{-1-\sqrt{5}}{2}$a(舍),
∴ b=$\frac{\sqrt{5}-1}{2}$a,
∴ AD=$\frac{\sqrt{5}-1}{2}$a,
∴ cos A = $\frac{AE}{AD}$ = $\frac{\frac{1}{2}a}{\frac{\sqrt{5}-1}{2}a}$ = $\frac{\sqrt{5}+1}{4}$.故选 B.

14. 新趋势 项目式学习 (济宁中考)知识再现:如图①,在$Rt△ ABC$中,$∠ C=90°$,$∠ A$,$∠ B$,$∠ C$的对边分别为$a$,$b$,$c$.
$\because \sin A=\frac{a}{c}$,$\sin B=\frac{b}{c}$,$\therefore c=\frac{a}{\sin A}$,$c=\frac{b}{\sin B}$.
$\therefore \frac{a}{\sin A}=\frac{b}{\sin B}$.

(1)拓展探究:如图②,在锐角三角形$ABC$中,$∠ A$,$∠ B$,$∠ C$的对边分别为$a$,$b$,$c$.请探究$\frac{a}{\sin A}$,$\frac{b}{\sin B}$,$\frac{c}{\sin C}$之间的关系,并写出探究过程.
(2)解决问题:如图③,为测量点$A$到河对岸点$B$的距离,选取与点$A$在河岸同一侧的点$C$,测得$AC=60\ \mathrm{m}$,$∠ A=75°$,$∠ C=60°$.请用拓展探究中的结论,求点$A$到点$B$的距离.

(1)作CD⊥AB于点D,AE⊥BC于点E.在Rt△ABE中,sin B = $\frac{AE}{AB}$ = $\frac{AE}{c}$,同理,sin B = $\frac{CD}{BC}$ = $\frac{CD}{a}$,sin∠BAC = $\frac{CD}{AC}$ = $\frac{CD}{b}$,sin∠BCA = $\frac{AE}{AC}$ = $\frac{AE}{b}$,
∴ AE = c sin B,AE = b sin∠BCA,CD = a sin B,CD = b sin∠BAC,
∴ c sin B = b sin∠BCA, a sin B = b sin∠BAC,
∴ $\frac{b}{\sin B}$ = $\frac{c}{\sin∠BCA}$, $\frac{a}{\sin∠BAC}$ = $\frac{b}{\sin B}$,
∴ $\frac{a}{\sin∠BAC}$ = $\frac{b}{\sin B}$ = $\frac{c}{\sin∠BCA}$,即 $\frac{a}{\sin A}$ = $\frac{b}{\sin B}$ = $\frac{c}{\sin C}$.
一题多解 如图①,作三角形ABC的外接圆,O为圆心,连接BO并延长交圆于点D,连接CD.设BD=2R.则根据直径所对的圆周角是直角以及同弧所对的圆周角相等可以得∠BCD=90°,∠A=∠D,
∴ sin A = sin D = $\frac{a}{2R}$,
∴ $\frac{a}{\sin A}$ = 2R.同理可得 $\frac{b}{\sin B}$ = 2R, $\frac{c}{\sin C}$ = 2R,
∴ $\frac{a}{\sin A}$ = $\frac{b}{\sin B}$ = $\frac{c}{\sin C}$.
(2) 如图②,过点B作BD⊥AC交AC于点D,过点A作AE⊥BC交BC于点E,在△ABC中,∠CBA=180°-∠BAC-∠C=180°-75°-60°=45°.
∵ ∠C=60°,
∴ ∠CBD=30°,
∴ BC=2CD,
∴ BD=$\sqrt{3}$ CD.
∵ ∠CBA=45°,
∴ ∠BAE=45°,
∴ BE=AE,AB=$\sqrt{2}$ AE.
∴ sin C = $\frac{BD}{BC}$ = $\frac{\sqrt{3} CD}{2CD}$ = $\frac{\sqrt{3}}{2}$, sin∠CBA = $\frac{AE}{AB}$ = $\frac{AE}{\sqrt{2} AE}$ = $\frac{\sqrt{2}}{2}$.
∵ $\frac{AB}{\sin C}$ = $\frac{AC}{\sin∠CBA}$,
∴ $\frac{AB}{\frac{\sqrt{3}}{2}}$ = $\frac{60}{\frac{\sqrt{2}}{2}}$,解得AB=30$\sqrt{6}$,故点A到点B的距离为30$\sqrt{6}$ m.