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2025年1加1轻巧夺冠完美期末七年级数学上册北师大版辽宁专版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年1加1轻巧夺冠完美期末七年级数学上册北师大版辽宁专版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
23.（13分）
如图，$O$为直线$EF$上一点，将一副直角三角尺的两个直角顶点叠合在$O$处，其中一个直角三角尺的另一顶点落在直线$EF$上的点$B$处.（本题涉及的角均小于或等于$180^{\circ}$）
（1）如图1，若$\angle AOC = 30^{\circ}$，求$\angle DOB$的度数.
（2）将直角三角尺$ODC$从图1的位置绕点$O$按逆时针方向旋转$\alpha(0^{\circ} ＜ \alpha ＜ 90^{\circ})$. 若$\angle AOD = 5\angle AOC$，求$\angle DOB$的度数.
（3）将直角三角尺$ODC$从如图2的位置绕点$O$按逆时针方向旋转一周，射线$ON$在$\angle AOD$内，射线$OM$在$\angle AOC$内，且$\angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，在转动过程中某个位置测得$\angle MON = 63^{\circ}$，请直
如图，$O$为直线$EF$上一点，将一副直角三角尺的两个直角顶点叠合在$O$处，其中一个直角三角尺的另一顶点落在直线$EF$上的点$B$处.（本题涉及的角均小于或等于$180^{\circ}$）
（1）如图1，若$\angle AOC = 30^{\circ}$，求$\angle DOB$的度数.
（2）将直角三角尺$ODC$从图1的位置绕点$O$按逆时针方向旋转$\alpha(0^{\circ} ＜ \alpha ＜ 90^{\circ})$. 若$\angle AOD = 5\angle AOC$，求$\angle DOB$的度数.
（3）将直角三角尺$ODC$从如图2的位置绕点$O$按逆时针方向旋转一周，射线$ON$在$\angle AOD$内，射线$OM$在$\angle AOC$内，且$\angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，在转动过程中某个位置测得$\angle MON = 63^{\circ}$，请直
接
写
出
$n$的值.
答案：
23.解：
(1)由题意得$\angle AOB = \angle COD = 90^{\circ}$，$\because \angle AOC = 30^{\circ}$，$\therefore \angle AOD = \angle COD - \angle AOC = 90^{\circ} - 30^{\circ} = 60^{\circ}$，$\therefore \angle DOB = \angle AOB + \angle AOD = 90^{\circ} + 60^{\circ} = 150^{\circ}$.
(2)分两种情况：①当OC在OA的右侧时，如答图1，$\because \angle AOD = 5\angle AOC$，$\because \angle AOD + \angle AOC = 90^{\circ}$，$\therefore \angle AOD = 90^{\circ} × \frac{5}{6} = 75^{\circ}$，$\therefore \angle DOB = \angle AOD + \angle AOB = 75^{\circ} + 90^{\circ} = 165^{\circ}$；②当OC在OA的左侧时，如答图2，$\because \angle AOD = 5\angle AOC$，$\therefore \angle COD = 4\angle AOC$，$\because \angle COD = 90^{\circ}$，$\therefore \angle AOC = \frac{1}{4}\angle COD = \frac{1}{4} × 90^{\circ} = 22.5^{\circ}$，$\therefore \angle DOB = 360^{\circ} - (\angle COD + \angle AOC + \angle AOB) = 360^{\circ} - (90^{\circ} + 22.5^{\circ} + 90^{\circ}) = 157.5^{\circ}$.
综上所述，$\angle DOB$的度数为$165^{\circ}$或$157.5^{\circ}$.
(3)分四种情况：①当OD在$\angle AOE$的内部时，如答图3，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD + \angle AOC = n\angle AON + n\angle AOM = n\angle MON$，$\because \angle AOD + \angle AOC = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$；②当OC在$\angle AOE$的内部时，如答图4，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD - \angle AOC = n\angle AON - n\angle AOM = n\angle MON$，$\because \angle AOD - \angle AOC = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$；③当OC，OD均在EF的下方时，如答图5，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD + \angle AOC = n\angle AON + n\angle AOM = n\angle MON$，$\because \angle AOD + \angle AOC = 360^{\circ} - \angle COD = 360^{\circ} - 90^{\circ} = 270^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 270 = 63n$，解得$n = \frac{30}{7}$；④当OD在$\angle AOB$的内部时，如答图6，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOC - \angle AOD = n\angle AOM - n\angle AON = n\angle MON$，$\because \angle AOC - \angle AOD = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$.
综上所述，$n$的值为$\frac{10}{7}$或$\frac{30}{7}$.
23.解：
(1)由题意得$\angle AOB = \angle COD = 90^{\circ}$，$\because \angle AOC = 30^{\circ}$，$\therefore \angle AOD = \angle COD - \angle AOC = 90^{\circ} - 30^{\circ} = 60^{\circ}$，$\therefore \angle DOB = \angle AOB + \angle AOD = 90^{\circ} + 60^{\circ} = 150^{\circ}$.
(2)分两种情况：①当OC在OA的右侧时，如答图1，$\because \angle AOD = 5\angle AOC$，$\because \angle AOD + \angle AOC = 90^{\circ}$，$\therefore \angle AOD = 90^{\circ} × \frac{5}{6} = 75^{\circ}$，$\therefore \angle DOB = \angle AOD + \angle AOB = 75^{\circ} + 90^{\circ} = 165^{\circ}$；②当OC在OA的左侧时，如答图2，$\because \angle AOD = 5\angle AOC$，$\therefore \angle COD = 4\angle AOC$，$\because \angle COD = 90^{\circ}$，$\therefore \angle AOC = \frac{1}{4}\angle COD = \frac{1}{4} × 90^{\circ} = 22.5^{\circ}$，$\therefore \angle DOB = 360^{\circ} - (\angle COD + \angle AOC + \angle AOB) = 360^{\circ} - (90^{\circ} + 22.5^{\circ} + 90^{\circ}) = 157.5^{\circ}$.
综上所述，$\angle DOB$的度数为$165^{\circ}$或$157.5^{\circ}$.
(3)分四种情况：①当OD在$\angle AOE$的内部时，如答图3，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD + \angle AOC = n\angle AON + n\angle AOM = n\angle MON$，$\because \angle AOD + \angle AOC = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$；②当OC在$\angle AOE$的内部时，如答图4，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD - \angle AOC = n\angle AON - n\angle AOM = n\angle MON$，$\because \angle AOD - \angle AOC = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$；③当OC，OD均在EF的下方时，如答图5，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOD + \angle AOC = n\angle AON + n\angle AOM = n\angle MON$，$\because \angle AOD + \angle AOC = 360^{\circ} - \angle COD = 360^{\circ} - 90^{\circ} = 270^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 270 = 63n$，解得$n = \frac{30}{7}$；④当OD在$\angle AOB$的内部时，如答图6，$\because \angle AOD = n\angle AON$，$\angle AOC = n\angle AOM$，$\therefore \angle AOC - \angle AOD = n\angle AOM - n\angle AON = n\angle MON$，$\because \angle AOC - \angle AOD = \angle COD = 90^{\circ}$，$\angle MON = 63^{\circ}$，$\therefore 90 = 63n$，解得$n = \frac{10}{7}$.
综上所述，$n$的值为$\frac{10}{7}$或$\frac{30}{7}$.
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