作DH⊥CE，则DH=CO=OB

∵CE⊥AB

∴DH∥AB

∴ $\frac{O E}{E D} = \frac{O F}{D H}$ =x，即OF=OBx

∴CF= $\sqrt{O C^{2} - O F^{2}}$ = $\sqrt{1 - x^{2}}$ OB

∴在Rt△CBF中，tanB= $\frac{C F}{C B} = \frac{\sqrt{1 - x^{2}} O B}{(1 + x) O B} = \frac{\sqrt{1 - x^{2}}}{(1 + x)}$ ----12分

②∵△COD的面积是△BOD的面积的3倍

∴ $\frac{C D}{C B} = \frac{3}{4}$

∵DH∥AB

∴ $\frac{D H}{B F} = \frac{C D}{C B} = \frac{3}{4}$

由DH=OB，得OF= $\frac{1}{3}$ DH

∴ $\frac{O E}{E D} = \frac{O F}{D H} = \frac{1}{3}$

∴y= $\frac{\sqrt{1 - {(\frac{1}{3})}^{2}}}{1 + \frac{1}{3}} = \frac{\sqrt{2}}{2}$

浙江省宁波市南三县2020年初中学业水平模拟考试数学试卷