题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201104/5/9146174a.png)
150
度.分析:首先由OA=OB=OC,得出∠OAB+∠OBA+∠OBC+∠OCB=140°,进而由四边形内角和定理得出,∠OAB+∠ABC+∠OCB=140°,再利用四边形内角和从而可得出答案.
解答:
解法1:∵OA=OB=OC,
∴∠OAB=∠OBA,∠OBC=∠OCB,
∵∠ABC=∠OBA+∠OBC=70°,
∴∠OAB+∠OBA+∠OBC+∠OCB=140°,即∠OAB+∠ABC+∠OCB=140°,
又∵∠ABC+∠BCD+∠ADC+∠BAD=360°,即∠ABC+∠OCB+∠OCD+∠ADC+∠DAO+∠OAB=360°,∵∠ADC=70°,∠OAB+∠ABC+∠OCB=140°,
∴∠DAO+∠DCO=360°-140°-70°=150°.
解法2:由AO=BO=CO,可知O是三角形ABC的外心,
∠ABC是圆周角,∠AOC是圆心角,
所以∠AOC=2∠ABC=140°,
又∠D=70°,
所以∠DAO+∠DCO=360°-140°-70°=150°.
答案为:150.
![](http://thumb.1010pic.com/pic3/upload/images/201104/5/9146174a.png)
∴∠OAB=∠OBA,∠OBC=∠OCB,
∵∠ABC=∠OBA+∠OBC=70°,
∴∠OAB+∠OBA+∠OBC+∠OCB=140°,即∠OAB+∠ABC+∠OCB=140°,
又∵∠ABC+∠BCD+∠ADC+∠BAD=360°,即∠ABC+∠OCB+∠OCD+∠ADC+∠DAO+∠OAB=360°,∵∠ADC=70°,∠OAB+∠ABC+∠OCB=140°,
∴∠DAO+∠DCO=360°-140°-70°=150°.
解法2:由AO=BO=CO,可知O是三角形ABC的外心,
∠ABC是圆周角,∠AOC是圆心角,
所以∠AOC=2∠ABC=140°,
又∠D=70°,
所以∠DAO+∠DCO=360°-140°-70°=150°.
答案为:150.
点评:此题主要考查了多边形内角和定理,以及等腰三角形的性质,解决问题的关键是得出∠OAB+∠ABC+∠OCB=140°,进而求出∠OAB+∠ABC+∠OCB=140°是解决问题的关键.
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