【答案】
分析:①连接OA,OC,过O作OD⊥AC于D,求出CD、AD,由勾股定理求出OD,求出∠ACO推出∠AOC=120°,根据圆周角定理求出∠B=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/0.png)
∠AOC,代入求出即可.②同样可求出∠D=60°,根据圆内接四边形性质求出∠ABC=120°.
解答:解:如图1,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/images1.png)
连接OA,OC,过O作OD⊥AC于D,
∵OD⊥AC,OD过圆心O,
∴AD=CD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/1.png)
AC=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/2.png)
,
由勾股定理得:OD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/3.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/4.png)
=3,
即OD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/5.png)
OC,
∴∠DCO=30°,∠COD=60°,
同理∠AOD=60°,
∵∠B=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/6.png)
∠AOC,
∴∠B=60°.
②如图2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/images8.png)
∵由垂径定理得CM═3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201533164426981/SYS201311032015331644269022_DA/7.png)
,OC=6,由勾股定理得:OM=3,
∴∠OCM=30°,∴∠MOC=60°,
∴∠AOC=2∠MOC=120°,
由圆周角定理得:∠D=60°,
∵A、D、C、B四点共圆,
∴∠ABC=120°,
故答案为:60°或120°.
点评:本题考查了垂径定理,圆周角定理,勾股定理,含30度角的直角三角形,三角形的外接圆等知识点的应用,关键是求出∠AOC的度数,通过做此题培养了学生的推理能力和计算能力.