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已知∠AOB与∠COD互余(∠COD的两边不在∠AOB的内部),OM平分∠AOC,ON平分∠BOD,将∠COD绕着点O逆时针旋转,使∠BOC=α(0°≤α<180°).(1)若∠AOB=60°,∠COD=30°.①当α=0°时,即OB与OC重

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已知∠AOB与∠COD互余(∠COD的两边不在∠AOB的内部),OM平分∠AOC,ON平分∠BOD,将∠COD绕着点O逆时针旋转,使∠BOC=α(0°≤α<180°).
(1)若∠AOB=60°,∠COD=30°.
①当α=0°时,即OB与OC重合时,如图1,则∠MON=______.
②当α=90°时,即OA与OD在一条直线上,如图2,求∠MON的度数.
③当α=140°时,请补全图形(如图3),并求出∠MON的度数.
(2)若∠AOB=β,∠COD=γ(β>γ),则∠MON=______.
▼优质解答
答案和解析
(1)①∠MON=
1
2
∠AOC+
1
2
∠BOD=45°.
②当α=90°时,
∠MON=180°-(
1
2
∠AOC+
1
2
∠BOD)
=180°-[
1
2
(∠AOB+∠BOC)+
1
2
(∠COD+∠BOC)]
=180°-[
1
2
(60°+90°)+
1
2
(30°+90°)]
=45°.
③当α=140°时,
∵∠AOD=360°-60°-30°-140°=130°,
∴∠MON=
1
2
∠AOC+
1
2
∠BOD-∠COD
=
1
2
(∠AOD+∠DOC)+
1
2
(∠BOC+∠COD)-∠COD
=
1
2
(∠AOD+∠BOC)
=
1
2
(360°-90°)
=135°;

(2)当∠AOB=β,∠COD=γ(β>γ)时,∠AOB与∠COD互余,则β+γ=90°,
当如图1所示:∠MON=
1
2
∠AOC+
1
2
∠BOD=
1
2
(β+γ)=45°,
如图3所示:
∠MON=
1
2
∠AOC+
1
2
∠BOD-∠COD
=
1
2
(∠AOD+∠DOC)+
1
2
(∠BOC+∠COD)-∠COD
=
1
2
(∠AOD+∠BOC)
=
1
2
(360°-∠AOB-∠COD)
=
1
2
(360°-90°)
=135°,
则∠MON=135°或45°.
故答案为:135°或45°.