《2025年课时夺冠九年级数学上册湘教版》

12. 已知$∠A$为锐角,且满足$|\sqrt {3}-2sinA|= 0$,则$cosA$的值为()
A. $\frac {1}{2}$
B. $\frac {\sqrt {2}}{2}$
C. $\frac {\sqrt {3}}{2}$
D. 1

答案： A

13. 在等腰三角形$ABC$中,$AB= AC= 10$,$BC= 12$,则$cos\frac {A}{2}$的值是()
A. $\frac {3}{5}$
B. $\frac {4}{5}$
C. $\frac {3}{4}$
D. $\frac {5}{4}$

答案： B

14. 如图,在$△ABC$中,$∠C= 90^{\circ }$,点$D在BC$上,$BD= 6$,$AD= BC$,$cos∠ADC= \frac {3}{5}$,则$CD$的长为()

A. 9
B. 15
C. 3
D. 6

答案： A

15. 如图,小正方形的边长均为1,点$A$,$B$,$C$都在格点上,则$∠ABC$的余弦值是______。

答案： $\frac{2\sqrt{5}}{5}$

16. 如图,在$//ogram ABCD$中,$AB⊥AC$。已知$AB= 10\sqrt {6}$,$//ogram ABCD的面积为200\sqrt {3}$。
(1)求$cosD$的值；
(2)求$∠BAD$的度数。

答案： 解：（1）$\because AB \perp AC$，$\therefore S_{\square ABCD} = AB \cdot AC = 10\sqrt{6}AC = 200\sqrt{3}$，$\therefore AC = 10\sqrt{2}$，$\therefore BC = \sqrt{AB^{2} + AC^{2}} = 20\sqrt{2}$。$\because$ 四边形$ABCD$是平行四边形，$\therefore \angle B = \angle D$，$\therefore \cos D = \cos B = \frac{AB}{BC} = \frac{\sqrt{3}}{2}$。（2）由（1）知$\cos D = \frac{\sqrt{3}}{2}$，$\therefore \angle D = 30^{\circ}$。$\because$ 四边形$ABCD$是平行四边形，$\therefore AB // CD$，$\therefore \angle BAD = 180^{\circ} - \angle D = 150^{\circ}$。

17. (永州新田县期末)如图,在平面直角坐标系中,$O$为原点,点$A的坐标为(10,0)$,点$B$在第一象限,$BO= 5$,$cos∠BOA= \frac {4}{5}$。
(1)求点$B$的坐标；
(2)求$cos∠BAO$的值。

答案： 解：（1）过点$B$作$BH \perp OA$于点$H$，$\therefore \angle BHO = 90^{\circ}$。在$Rt\triangle BOH$中，$\cos \angle BOA = \frac{OH}{OB} = \frac{4}{5}$，$\therefore \frac{OH}{5} = \frac{4}{5}$，$\therefore OH = 4$，$\therefore BH = \sqrt{OB^{2} - OH^{2}} = 3$，$\therefore$ 点$B$的坐标为$(4, 3)$。（2）由（1）知$OH = 4$，$BH = 3$。$\because$ 点$A$的坐标为$(10, 0)$，$\therefore OA = 10$，$\therefore AH = OA - OH = 6$。$\because BH \perp OA$，$\therefore \angle BHA = 90^{\circ}$，$\therefore$ 在$Rt\triangle BHA$中，$AB = \sqrt{BH^{2} + AH^{2}} = 3\sqrt{5}$，$\therefore \cos \angle BAO = \frac{AH}{AB} = \frac{6}{3\sqrt{5}} = \frac{2\sqrt{5}}{5}$。

18. (张家界永定区期末)如图,在四边形$ABCD$中,$∠ABC= 90^{\circ }$,$AD= BD$,$BC= CD$。
(1)若$BD= 13$,$AB= 10$,求$cos∠CBD$的值；
(2)设$△ABD的面积为S_{1}$,$△BCD的面积为S_{2}$,求证：$\frac {S_{1}}{S_{2}}= 4cos^{2}∠CBD$。

答案： （1）解：过点$D$作$DE \perp AB$于点$E$，$\therefore \angle AED = 90^{\circ} = \angle ABC$，$\therefore DE // BC$，$\therefore \angle CBD = \angle BDE$。$\because BD = AD$，$\therefore BE = AE = \frac{1}{2}AB = 5$。在$Rt\triangle BED$中，$DE = \sqrt{BD^{2} - BE^{2}} = \sqrt{13^{2} - 5^{2}} = 12$，$\therefore \cos \angle CBD = \cos \angle BDE = \frac{DE}{DB} = \frac{12}{13}$。
（2）证明：过点$C$作$CF \perp BD$于点$F$，$\therefore \angle BFC = 90^{\circ}$。$\because DE \perp AB$，$\therefore \angle BED = 90^{\circ} = \angle BFC$。$\because BC = CD$，$\therefore BD = 2BF$。由（1）知$\angle CBD = \angle BDE$，$\therefore \triangle DEB \sim \triangle BFC$，$\therefore \frac{S_{1}}{S_{2}} = (\frac{DE}{BF})^{2} = 4 \times (\frac{DE}{2BF})^{2} = 4 \times (\frac{DE}{DB})^{2}$。由（1）知$\cos \angle CBD = \frac{DE}{DB}$，$\therefore \frac{S_{1}}{S_{2}} = 4\cos^{2} \angle CBD$。