【答案】
分析:由題意,半徑為1,弦AB、AC分別是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/0.png)
、
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/1.png)
,
作OM⊥AB,ON⊥AC;利用余弦函數(shù),可求出∠OAM=45°,∠OAN=30°;
AC的位置情況有兩種,如圖所示;故∠BAC的度數(shù)為45°+30°或45°-30°.問題可求.
解答:解:作OM⊥AB,ON⊥AC;由垂徑定理,可得AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/2.png)
,AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/3.png)
,
∵弦AB、AC分別是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/4.png)
、
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/5.png)
,∴AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/6.png)
,AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/7.png)
;
∵半徑為1∴OA=1;
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/9.png)
∴∠OAM=45°;同理,∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/11.png)
,∴∠OAN=30°;
∴∠BAC=∠OAM+∠OAN或∠OAM-∠OAN
∴∠BAC=75°或15°.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001141233133599/SYS201311030011412331335013_DA/images12.png)
點評:本題綜合性強(qiáng),關(guān)鍵是畫出圖形,作好輔助線,利用垂徑定理和直角三角形的特殊余弦值求得角的度數(shù).