如图,直线AB与直线CD相交于点O,E是∠AOD内一点,已知OE⊥AB,∠BOD=45°,则∠COE的度数是______.
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如图,直线AB与直线CD相交于点O,E是∠AOD内一点,已知OE⊥AB,∠BOD=45°,则∠COE的度数是______.
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174913-87317.png) |
答案
∵OE⊥AB, ∴∠AOE=90°, ∵∠BOD=45°, ∴∠AOC=∠BOD=45°, ∴∠COE=∠AOE+∠AOC=90°+45°=135°. 故答案为:135°. |
举一反三
如图,已知OE是∠BOC的平分线,OD是∠AOC的平分线,且∠AOB=150°,则∠DOE的度数是______度.
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174908-47587.png) |
如图,已知∠DOC=42°,OD平分∠AOC,∠BOD=14°,求∠AOB的度数.
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174856-80106.png) |
如图,OA⊥OB,∠1=∠2,∠3=∠4,则∠EOF=( )
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174849-93077.png) |
如图所示,两块三角板的直角顶点O重叠在一起,且OB恰好平分∠COD,则∠AOD的度数是______度.
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174844-46756.png) |
按图填空: (1)∠AOB+∠BOC=______; (2)∠AOC+∠COD=______; (3)∠BOD-∠COD=______; (4)∠AOD-∠BOD=______; (5)∠AOD=∠AOB+∠BOC+______=∠AOB+______=∠AOC+______; (6)∠BOC=∠AOD-∠AOB-______=∠AOC-______=∠BOD-______⋅
![](http://img.shitiku.com.cn/uploads/allimg/20191118/20191118174831-43202.png) |
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