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1. [2025常州中考]如图，在$\mathrm{Rt}△ ABC$中，$∠ A = 90°$，$AB = 3$，$AC = 4$，则$\sin B$的值是（
A.$\frac{3}{5}$
B.$\frac{3}{4}$
C.$\frac{4}{5}$
D.$\frac{4}{3}$
C
）A.$\frac{3}{5}$
B.$\frac{3}{4}$
C.$\frac{4}{5}$
D.$\frac{4}{3}$
答案
1 C 在$\mathrm{Rt}△ABC$中，由勾股定理得$BC = \sqrt{AB^2 + AC^2} = \sqrt{3^2 + 4^2} = 5$，$\therefore \sin B = \frac{AC}{BC} = \frac{4}{5}$.
2. [2026西交大苏州附中段考]第七届国际数学教育大会(ICME)的会徽如图1所示,在其主体图案中选择两个相邻的直角三角形,恰好能够组合得到如图2所示的四边形OABC.若$OC=\sqrt{5}$,$AB=BC=1$,则$\sin∠ OBA$的值为 (
A.$\frac{\sqrt{3}}{3}$
B.$\frac{\sqrt{3}}{2}$
C.$\frac{1}{3}$
D.$\frac{1}{2}$
B
)A.$\frac{\sqrt{3}}{3}$
B.$\frac{\sqrt{3}}{2}$
C.$\frac{1}{3}$
D.$\frac{1}{2}$
答案
2 B 由题意，得$∠A = ∠OBC = 90°$. $\because OC = \sqrt{5}, BC = 1$，$\therefore OB = \sqrt{OC^2 - BC^2} = 2$. 同理，得$AO = \sqrt{3}$，$\therefore \sin∠OBA = \frac{AO}{BO} = \frac{\sqrt{3}}{2}$.
3. 教材P130T2变式 [2026安庆段考] 如图，在$△ ABC$中，$∠ ACB=90°$，$CD⊥ AB$于点$D$，若$CD:AC=3:4$，则$\sin∠ BCD$的值是（
A.$\dfrac{3}{4}$
B.$\dfrac{3}{5}$
C.$\dfrac{4}{5}$
D.$\dfrac{9}{34}$
A
）A.$\dfrac{3}{4}$
B.$\dfrac{3}{5}$
C.$\dfrac{4}{5}$
D.$\dfrac{9}{34}$
答案
3 A 解题思路：
由题意可知$∠BCD + ∠B = ∠B + ∠A = 90°$，$\therefore ∠BCD = ∠A$.
$\because CD:AC = 3:4$，$\therefore \sin∠BCD = \sin A = \frac{CD}{AC} = \frac{3}{4}$(等角转化).
提分大招
在三角函数相关问题中，通常借助余角(或补角)、平行线、等腰三角形、全等三角形、相似三角形等知识来寻找等角关系.
4. 教材P130T3 变式 [2024 临夏州中考] 如图,在△ABC中,AB=AC=5,sin B=$\frac{4}{5}$,则BC的长是 (
A.3
B.6
C.8
D.9
B
)A.3
B.6
C.8
D.9
答案
4 B 如图,过点A作BC的垂线,垂足为M(构造直角三角形),则$\sin B = \frac{AM}{AB}$,$\therefore AM = AB · \sin B = 5 × \frac{4}{5} = 4$,$\therefore BM = \sqrt{AB^2 - AM^2} = \sqrt{5^2 - 4^2} = 3$. 又$\because AB = AC$,$\therefore BC = 2BM = 6$.
得分要点
已知条件中有等腰三角形时,常作底边上的高构造直角三角形,根据等腰三角形“三线合一”的性质可知这条高也平分底边.
5. [2026 合肥瑶海区段考] 在$△ ABC$中，$∠ C = 90°$，$BC = 8$，$AB = 17$，则$\cos A$的值是（
A.$\dfrac{15}{17}$
B.$\dfrac{8}{17}$
C.$\dfrac{8}{15}$
D.$\dfrac{15}{8}$
A
）A.$\dfrac{15}{17}$
B.$\dfrac{8}{17}$
C.$\dfrac{8}{15}$
D.$\dfrac{15}{8}$
答案
5 A $\because ∠C = 90°, BC = 8, AB = 17$,$\therefore AC = \sqrt{AB^2 - BC^2} = \sqrt{17^2 - 8^2} = 15$,$\therefore \cos A = \frac{AC}{AB} = \frac{15}{17}$.
6. 在$\mathrm{Rt}△ ABC$中，若$∠ C = 90°$，$BC = 2AC$，则$\cos A$的值为（
A.$\frac{1}{2}$
B.$\frac{\sqrt{3}}{2}$
C.$\frac{2\sqrt{5}}{5}$
D.$\frac{\sqrt{5}}{5}$
D
）A.$\frac{1}{2}$
B.$\frac{\sqrt{3}}{2}$
C.$\frac{2\sqrt{5}}{5}$
D.$\frac{\sqrt{5}}{5}$
答案
6 D 设$AC = k(k ＞ 0)$(设参法),则$BC = 2k$,由勾股定理得,$AB = \sqrt{5}k$,故$\cos A = \frac{AC}{AB} = \frac{\sqrt{5}}{5}$.
7. [2026合肥四十六中月考]如图，直线$y=\frac{3}{4}x + 3$与x，y轴分别交于A，B两点，则$\cos∠ BAO$的值是（
A.$\frac{4}{5}$
B.$\frac{3}{5}$
C.$\frac{4}{3}$
D.$\frac{5}{4}$
A
）A.$\frac{4}{5}$
B.$\frac{3}{5}$
C.$\frac{4}{3}$
D.$\frac{5}{4}$
答案
7 A 当$x = 0$时,$y = 3$,当$y = 0$时,$x = -4$,$\therefore A(-4,0)$,$B(0,3)$,$\therefore OA = 4, OB = 3$,由勾股定理,得$AB = 5$(坐标系隐含“直角”这个条件),则$\cos∠BAO = \frac{OA}{AB} = \frac{4}{5}$.
8. [2026 六安金安区期中]如图,在$△ ABC$中,$∠ C = 90°,AC = 8\ \mathrm{cm},AB$的垂直平分线$MN$交$AC$于点$D$,连接$BD$,若$\cos∠ BDC = \dfrac{3}{5}$,则$BC$的长是
4 cm
.答案
8 4 cm $\because MN$垂直平分$AB$,$\therefore BD = AD$. 在$\mathrm{Rt}△BCD$中,$\cos∠BDC = \frac{CD}{BD} = \frac{CD}{AD} = \frac{CD}{AC - CD} = \frac{3}{5}$,$\therefore \frac{CD}{8 - CD} = \frac{3}{5}$,解得$CD = 3$,经检验$CD = 3$是方程的解且符合实际,则$BD = 5\ \mathrm{cm}$,由勾股定理,得$BC = 4\ \mathrm{cm}$.
9. [2025安庆期末]在$Rt△ ABC$中，$∠ C=90°$，$AB=5$，$AC=4$，下列四个选项中，正确的是（
A.$\tan B=\dfrac{3}{4}$
B.$\tan A=\dfrac{4}{3}$
C.$\sin B=\dfrac{4}{5}$
D.$\cos B=\dfrac{4}{5}$
C
）A.$\tan B=\dfrac{3}{4}$
B.$\tan A=\dfrac{4}{3}$
C.$\sin B=\dfrac{4}{5}$
D.$\cos B=\dfrac{4}{5}$
答案
9 C 画草图,如图,由勾股定理,得$BC = 3$. 故$\tan B = \frac{AC}{BC} = \frac{4}{3}$,$\tan A = \frac{BC}{AC} = \frac{3}{4}$,$\sin B = \frac{AC}{AB} = \frac{4}{5}$,$\cos B = \frac{BC}{AB} = \frac{3}{5}$. 故C选项正确.
10. [2026亳州段考]在$△ ABC$中，$∠ C = 90°$，$\tan A = \dfrac{1}{3}$，则$\cos B =$
$\frac{\sqrt{10}}{10}$
答案
10 $\frac{\sqrt{10}}{10}$ 解题思路：
如图,在$△ABC$中,$∠C = 90°$,$\tan A = \frac{1}{3}$,$\therefore \tan A = \frac{BC}{AC} = \frac{1}{3}$,$\therefore$ 设$BC = a, AC = 3a$,由勾股定理得,$AB = \sqrt{BC^2 + AC^2} = \sqrt{a^2 + (3a)^2} = \sqrt{10}a$,$\therefore \cos B = \frac{BC}{AB} = \frac{a}{\sqrt{10}a} = \frac{\sqrt{10}}{10}$.
得分要点
在直角三角形中,给出某一锐角的一个三角函数值,求另一个三角函数值时,先利用锐角三角函数的概念,设出参数,结合勾股定理,用含参数的式子分别表示出直角三角形的三边长,再利用概念求解.
11. [2025西安雁塔区期中] 在$\mathrm{Rt}△ ABC$中，$∠ BCA = 90°$，$CD$是中线，$BC = 8$，$CD = 5$. 求$\sin∠ ACD$，$\cos∠ ACD$和$\tan∠ ACD$.
答案
11 解:画草图,如图,在$\mathrm{Rt}△ABC$中,$∠BCA = 90°$,$CD$是中线,$CD = 5$,
$\therefore AD = CD, AB = 2CD = 10$,$\therefore ∠A = ∠ACD$.
在$\mathrm{Rt}△ABC$中,$BC = 8$,$AB = 10$,
$\therefore AC = \sqrt{AB^2 - BC^2} = \sqrt{10^2 - 8^2} = 6$,
$\therefore \sin∠ACD = \sin A = \frac{BC}{AB} = \frac{8}{10} = \frac{4}{5}$(等角转化),
$\cos∠ACD = \cos A = \frac{AC}{AB} = \frac{6}{10} = \frac{3}{5}$,
$\tan∠ACD = \tan A = \frac{BC}{AC} = \frac{8}{6} = \frac{4}{3}$.
