《2025年核心素养新讲堂七年级数学》

12. 如图①，已知$\angle AOB = 120^{\circ}$，$\angle COD = 60^{\circ}$，$OM$ 在$\angle AOC$ 内，$ON$ 在$\angle BOD$ 内，$\angle AOM = \frac{1}{3}\angle AOC$，$\angle BON = \frac{1}{3}\angle BOD$。
(1)$\angle COD$ 从图①中的位置绕点 $O$ 逆时针旋转到 $OC$ 与 $OB$ 重合时，如图②，$\angle MON = $100$^{\circ}$；
(2)$\angle COD$ 从图②中的位置绕点 $O$ 逆时针旋转 $n^{\circ}(0 ＜ n ＜ 120$ 且 $n \neq 60)$，求$\angle MON$ 的度数；
(3)$\angle COD$ 从图②中的位置绕点 $O$ 顺时针旋转 $n^{\circ}(0 ＜ n ＜ 120)$，则 $n = $50或70 时，$\angle MON = 2\angle BOC$。

答案：
12.
(1)$100$
(2)分两种情况:
①当$0 ＜ n ＜ 60$时,如图①,
　　　　
设$\angle AOM = x^{\circ}$,则$\angle AOC = 3x^{\circ}$,$\angle MOC = \angle AOC - \angle AOM = 2x^{\circ}$,$\angle BOC = \angle AOB - \angle AOC = 120^{\circ} - 3x^{\circ}$.$\therefore \angle BOD = \angle COD - \angle BOC = 60^{\circ} - (120^{\circ} - 3x^{\circ}) = 3x^{\circ} - 60^{\circ}$,$\therefore \angle BON = \frac{1}{3}\angle BOD = x^{\circ} - 20^{\circ}$,$\angle MON = \angle MOC + \angle BOC + \angle BON = 2x^{\circ} + 120^{\circ} - 3x^{\circ} + x^{\circ} - 20^{\circ} = 100^{\circ}$.
②当$60 ＜ n ＜ 120$时,如图②,
　　　　　　
设$\angle AOM = x^{\circ}$,则$\angle AOC = 3x^{\circ}$,$\angle MOC = \angle AOC - \angle AOM = 2x^{\circ}$,$\therefore \angle BOD = \angle AOB - \angle COD - \angle AOC = 120^{\circ} - 60^{\circ} - 3x^{\circ} = 60^{\circ} - 3x^{\circ}$,$\therefore \angle DON = \frac{2}{3}\angle BOD = 40^{\circ} - 2x^{\circ}$,$\angle MON = \angle MOC + \angle COD + \angle DON = 2x^{\circ} + 60^{\circ} + 40^{\circ} - 2x^{\circ} = 100^{\circ}$.
(3)$50$或$70$
当$0 ＜ n \leq 60$时,如图③,
 
此时$\angle MON = 100^{\circ} = 2\angle BOC = 2n^{\circ}$,$\therefore n = 50^{\circ}$.
当$60 ＜ n ＜ 120$时,如图④,

$\angle AOC = 360^{\circ} - \angle AOB - \angle BOC = 240^{\circ} - n^{\circ}$,$\therefore \angle COM = \frac{2}{3}\angle AOC = \frac{2}{3}(240^{\circ} - n^{\circ}) = 160^{\circ} - \frac{2}{3}n^{\circ}$,$\angle DON = \frac{2}{3}\angle BOD = \frac{2}{3}(60^{\circ} + n^{\circ}) = 40^{\circ} + \frac{2}{3}n^{\circ}$,$\therefore \angle MON = \angle DON + \angle COM - \angle COD = 40^{\circ} + \frac{2}{3}n^{\circ} + 160^{\circ} - \frac{2}{3}n^{\circ} - 60^{\circ} = 140^{\circ}$.$\therefore 2\angle BOC = 2n^{\circ}$,$\therefore n = 70^{\circ}$.
综上所述$n = 50$或$70$.