题目内容
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_ST/images0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_ST/1.png)
A.(0,2)
B.(0,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_ST/2.png)
C.(0,4)
D.(0,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_ST/3.png)
【答案】分析:首先连接AB,由圆周角定理可得∠ABO=60°,然后由三角函数的性质,求得OB的长,则可得点B的坐标.
解答:
解:连接AB,
∵∠ABO=∠OPA=60°,∠AOB=90°,
∴OB=
=
=4.
∴点B的坐标为:(0,4).
故选C.
点评:此题考查了圆周角定理以及三角函数的性质.此题难度不大,注意辅助线的作法,注意掌握数形结合思想的应用.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_DA/images0.png)
∵∠ABO=∠OPA=60°,∠AOB=90°,
∴OB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193024485492883/SYS201311011930244854928009_DA/1.png)
∴点B的坐标为:(0,4).
故选C.
点评:此题考查了圆周角定理以及三角函数的性质.此题难度不大,注意辅助线的作法,注意掌握数形结合思想的应用.
![](http://thumb2018.1010pic.com/images/loading.gif)
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