1. 在$\mathrm{Rt}△ ABC$中，$∠ C=90°$，$a,b,c$分别为$∠ A$，$∠ B$，$∠ C$的对边，下列关系中错误的是（）

A.$b = c·\cos B$
B.$b = a·\tan B$
C.$b = c·\sin B$
D.$a = b·\tan A$

1 A 如图,
∵ $\cos B = \frac{a}{c}$,$\therefore a = c · \cos B$,$\therefore$ 选项 A 符合题意;$\because \tan B = \frac{b}{a}$,$\therefore b = a · \tan B$,$\therefore$ 选项 B 不符合题意;$\because \sin B = \frac{b}{c}$,$\therefore b = c · \sin B$,$\therefore$ 选项 C 不符合题意;$\because \tan A = \frac{a}{b}$,$\therefore a = b · \tan A$,$\therefore$ 选项 D 不符合题意.

2. [2025合肥庐阳区三模] 如图,在$Rt△ ABC$中,$∠ C=90°$,点D在BC边上,若$AC=5$,$BD=7$,$\tan∠ ADC=1$,则$\sin B$为 ()


A.$\frac{5}{13}$
B.$\frac{7}{13}$
C.$\frac{5}{12}$
D.$\frac{7}{12}$

2 A 在$\mathrm{Rt}△ACD$中,$\tan ∠ ADC = 1$,即 $\tan ∠ ADC = \frac{AC}{CD} = 1$,$\therefore CD = AC = 5$,$\therefore BC = BD + CD = 7 + 5 = 12$. 在 $\mathrm{Rt}△ABC$ 中,由勾股定理,得 $AB = \sqrt{AC^2 + BC^2} = 13$,$\therefore \sin B = \frac{AC}{AB} = \frac{5}{13}$.

3. 如图,在$Rt△ ABC$中,$∠ C=90°$,$\sin A=\frac{1}{3}$,则 ()

A.$\cos A=\frac{1}{3}$
B.$\tan A=2\sqrt{2}$
C.$\cos B=\frac{2\sqrt{2}}{3}$
D.$\tan B=2\sqrt{2}$
拍批+错题本

3 D $\because \sin A = \frac{1}{3}$,$\therefore \frac{BC}{AB} = \frac{1}{3}$,$\therefore$ 可设 $BC = x(x ＞ 0)$,则$AB = 3x$,$\therefore AC = \sqrt{AB^2 - BC^2} = 2\sqrt{2}x$,$\therefore \cos A = \frac{2\sqrt{2}x}{3x} = \frac{2\sqrt{2}}{3}$,$\tan A = \frac{x}{2\sqrt{2}x} = \frac{\sqrt{2}}{4}$,$\cos B = \frac{x}{3x} = \frac{1}{3}$,$\tan B = \frac{2\sqrt{2}x}{x} = 2\sqrt{2}$. 故只有选项 D 符合题意.

4. 新趋势·数学文化 [2024 资阳中考]第 14 届国际数学教育大会(ICME -14)会标如图1所示，会标中心的图案来源于我国古代数学家赵爽的“弦图”.如图2所示的“弦图”是由四个全等的直角三角形(△ABE，△BCF，△CDG，△DAH)和一个小正方形EFGH拼成的大正方形ABCD.若EF:AH=1:3，则sin∠ABE = ()


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

4 C $\because EF: AH = 1: 3$,$\therefore$ 可设 $EF = x$,则 $AH = 3x$. $\because △ ABE ≌ △ DAH$,四边形 EFGH 为正方形,$\therefore AH = BE = 3x$,$EF = HE = x$,$\therefore AE = 4x$. 在 $\mathrm{Rt}△ABE$ 中, 由勾股定理, 得 $AB = \sqrt{AE^2 + BE^2} = 5x$,$\therefore \sin ∠ ABE = \frac{AE}{AB} = \frac{4x}{5x} = \frac{4}{5}$.

5. [2025亳州期末]如图,在平面直角坐标系中,点$P(m,12)$在第二象限内.若$OP$与$x$轴负半轴的夹角$α$的正切值为$\frac{4}{3}$,则$m$的值为.

5 -9 如图,过点 $P$ 作 $PA ⊥ x$ 轴于点 $A$,则 $AP = 12$,$OA = -m$. $\because \tan α = \frac{AP}{OA} = \frac{12}{-m} = \frac{4}{3}$,$\therefore -4m = 3 × 12$,解得 $m = -9$.