【答案】
分析:根据题意画出图形,由于△ABC的形状不能确定,故应分锐角三角形与钝角三角形两种情况进行讨论.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/images0.png)
解:当△ABC是锐角三角形时,如图1所示:过点O作OD⊥AC于点D,
∵AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/0.png)
cm,OC=6cm,
∴OD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/1.png)
AC=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/2.png)
cm,
∴sin∠COD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/3.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/4.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/5.png)
,
∴∠COD=60°
∴∠B=∠COD=60°;
当△ABC是锐角三角形时,如图2所示:过点O作OD⊥AC于点D,
同理可得∠COD=60°,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/6.png)
所对的圆心角=360°-2×60°=240°,
∴∠B=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202010656903709/SYS201311032020106569037009_DA/7.png)
×240°=120°.
故答案为:60°或120°.
点评:本题考查的是圆周角定理及垂径定理,解答此题时要注意分类讨论,不要漏解.