《2025年全科王同步课时练习九年级数学上册青岛版》

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

A.$\frac{\sqrt{10}}{10}$
B.$\frac{2}{3}$
C.$\frac{3}{4}$
D.$\frac{3\sqrt{10}}{10}$

答案： D

13. (2023·枣庄台儿庄区模拟)如图,在平面直角坐标系中,点A,B分别在x轴负半轴和y轴正半轴上,点C在OB上,OC : BC = 1 : 2,连接AC,过点O作OP // AB交AC的延长线于P. 若P(1,1),则tan∠OAP的值是 ( )

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

答案：
C[提示:如图,过点P作PQ⊥x轴于点Q,
∵OP//AB,
∴△OCP∽△BCA,
∴CP:AC = OC:BC = 1:2.
∵∠AOC = ∠AQP = 90°,
∴CO//PQ,
∴OQ:AO = CP:AC = 1:2.
∵P(1,1),
∴PQ = OQ = 1,
∴AO = 2,
∴tan∠OAP = $\frac{PQ}{AQ}$ = $\frac{1}{1 + 2}$ = $\frac{1}{3}$.]

14. (2023·济宁泗水县二模)如图,在四边形ABCD中,E,F分别是AB,AD的中点,若EF = 6,BC = 13,CD = 5,则tan C等于______.

答案：
$\frac{12}{5}$[提示:如图,连接BD,
∵E,F分别是AB,AD的中点,
∴EF//BD,EF = $\frac{1}{2}$BD,
∴BD = 12,又
∵BC = 13,CD = 5,
∴△BDC是直角三角形,
∴tanC = $\frac{BD}{CD}$ = $\frac{12}{5}$]

15. 如图,在Rt△ABC中,∠C = 90°,M是直角边AC上一点,MN ⊥ AB于点N,AN = 3,AM = 4,求cos B的值.

答案： 解:
∵∠C = 90°,MN⊥AB,
∴∠C = ∠ANM = 90°.又
∵∠A = ∠A,
∴△AMN∽△ABC,
∴$\frac{AC}{AB}$ = $\frac{AN}{AM}$ = $\frac{3}{4}$.设AC = 3x,AB = 4x,由勾股定理得BC = $\sqrt{AB^2 - AC^2}$ = $\sqrt{7}x$,在Rt△ABC中,cosB = $\frac{BC}{AB}$ = $\frac{\sqrt{7}x}{4x}$ = $\frac{\sqrt{7}}{4}$.

16. 如图,在△ABC中,∠ACB = 90°,CD ⊥ AB于D,AC = 8,AB = 10,求cos∠BCD的值.

答案： 解:
∵∠ACB = 90°,CD⊥AB,
∴∠BDC = ∠ACB = 90°,
∴∠B + ∠BCD = 90°,∠B + ∠A = 90°,
∴∠BCD = ∠A.
∵AB = 10,AC = 8,
∴cos∠BCD = cosA = $\frac{AC}{AB}$ = $\frac{8}{10}$ = $\frac{4}{5}$.

17. 如图,在4 × 4的正方形网格中,△ABC和△DEF的顶点都在边长为1的小正方形顶点上,求tan∠ACB的值.

答案： 解:由图知AB = EF = 2.由勾股定理,得BC = 2$\sqrt{2}$,AC = 2$\sqrt{5}$,DF = $\sqrt{10}$,DE = $\sqrt{2}$,
∴$\frac{FD}{AC}$ = $\frac{ED}{AB}$ = $\frac{FE}{BC}$ = $\frac{\sqrt{2}}{2}$
∴△FDE ∽△CAB,
∴∠DFE = ∠ACB,
∴tan∠ACB = tan∠DFE = $\frac{1}{3}$.

18. 如图,在Rt△ABC中,∠C = 90°,AC = 3,BC = 4,∠ABC的平分线交边AC于点D,延长BD至点E,且BD = 2DE,连接AE.
(1) 求线段CD的长；
(2) 求△ADE的面积.

答案：
解:
(1)如图,过点D作DH⊥AB,垂足为点H,
∵BD平分∠ABC,∠C = 90°,
∴设DH = DC = x,则AD = 3 - x.
∵∠C = 90°,AC = 3,BC = 4,
∴AB = 5.
∵sin∠BAC = $\frac{HD}{AD}$ = $\frac{BC}{AB}$,
∴$\frac{x}{3 - x}$ = $\frac{4}{5}$,
∴x = $\frac{4}{3}$,即CD = $\frac{4}{3}$.
(2)$S_{\triangle ABD}$ = $\frac{1}{2}$AB·DH = $\frac{1}{2}$×5×$\frac{4}{3}$ = $\frac{10}{3}$
∵BD = 2DE,
∴$\frac{S_{\triangle ABD}}{S_{\triangle ADE}}$ = $\frac{BD}{DE}$ = 2,
∴$S_{\triangle ADE}$ = $\frac{10}{3}$×$\frac{1}{2}$ = $\frac{5}{3}$.