Question

如图1，已知∠AOB=80°，∠COD==40°,OM平分∠BOD,ON平分∠AOC.

(1)将图1中∠COD绕O点逆时针旋转,使射线OC与
(2)将图2∠COD绕O点逆时针旋转α度,其他条件不变,
①当40°< α<100°，请完成图（3），并求∠MON的度数；
②当140°<α<180°，请完成图（4），并求∠MON的度数。

Answer 1

要分四种情况讨论:
(1)点C,D都在∠AOB内部时,
∠COD=∠AOC-∠AOD
=60°-40°
=20°;
(2)点C在∠AOB内部,点D在∠AOB外部时,
∠COD=∠BOD+∠BOC
=40°+20°
=60°
(3)点C在∠AOB外部,点D在∠AOB内部时,
∠COD=∠AOC+∠AOD
=60°+40°
=100°
(4)点C在∠AOB外部,点D也在∠AOB外部时
∠COD=∠AOC+∠AOB+∠BOD
=60°+80°+40°
=180°

Answer 2

60度
∠NOC=(∠AOB+∠BOC)/2
∠MOB=(∠COD+∠BOC)/2
∠MON=∠NOC+∠MOB-∠BOC
=(∠AOB+∠BOC)/2 + (∠COD+∠BOC)/2 -∠BOC
=(∠AOB+∠COD)/2
=(80+40)/2
=60
题2同解，还是60

Answer 3

因为∠NOC=(∠AOB+∠BOC)/2
∠MOB=(∠COD+∠BOC)/2
∠MON=∠NOC+∠MOB-∠BOC
=(∠AOB+∠BOC)/2 + (∠COD+∠BOC)/2 -∠BOC
=(∠AOB+∠COD)/2
=(80+40)/2
所以 =60
题2同解，都是60

Answer 4

60度
∠NOC=(∠AOB+∠BOC)/2
∠MOB=(∠COD+∠BOC)/2
∠MON=∠NOC+∠MOB-∠BOC
=(∠AOB+∠BOC)/2 + (∠COD+∠BOC)/2 -∠BOC
=(∠AOB+∠COD)/2
=(80+40)/2
=60

Answer 5

60度
∠NOC=(∠AOB+∠BOC)/2
∠MOB=(∠COD+∠BOC)/2
∠MON=∠NOC+∠MOB-∠BOC
=(∠AOB+∠BOC)/2 + (∠COD+∠BOC)/2 -∠BOC
=(∠AOB+∠COD)/2
=(80+40)/2
=60
题2同解，还是60