《2025年经纶学典5星学霸七年级数学上册苏科版》

1. (2025·南京期末)如图,直线$AB与直线CD相交于点O$,$OE平分\angle BOD$。
(1)若$\angle BOE = 41^{\circ}36'$,则$\angle COB= $$96^{\circ}48'$。
(2)若$OF平分\angle COE$,$\angle FOB的度数为30^{\circ}$。
①求$\angle AOC$的度数；
②作射线$OG\perp OE$,请直接写出$\angle FOG$的度数。

(2)①因为OE平分$∠BOD$，所以$∠EOD = ∠BOE$。设$∠EOD = ∠BOE = x^{\circ}$，因为$∠FOB = 30^{\circ}$，所以$∠FOE = ∠FOB + ∠BOE = (30 + x)^{\circ}$。因为OF平分$∠COE$，所以$∠COE = 2∠FOE = (60 + 2x)^{\circ}$。因为$∠COE + ∠EOD = 180^{\circ}$，所以$60 + 2x + x = 180$，解得$x = 40$，所以$∠BOD = 2×40^{\circ} = 80^{\circ}$，所以$∠AOC = ∠BOD = 80^{\circ}$。
②$20^{\circ}$或$160^{\circ}$

答案： 1.
(1)$96^{\circ}48'$
(2)①因为OE平分$∠BOD$，所以$∠EOD = ∠BOE$。设$∠EOD = ∠BOE = x^{\circ}$，因为$∠FOB = 30^{\circ}$，所以$∠FOE = ∠FOB + ∠BOE = (30 + x)^{\circ}$。因为OF平分$∠COE$，所以$∠COE = 2∠FOE = (60 + 2x)^{\circ}$。因为$∠COE + ∠EOD = 180^{\circ}$，所以$60 + 2x + x = 180$，解得$x = 40$，所以$∠BOD = 2×40^{\circ} = 80^{\circ}$，所以$∠AOC = ∠BOD = 80^{\circ}$。
②由①知$∠FOE = ∠FOB + ∠BOE = 30^{\circ} + 40^{\circ} = 70^{\circ}$。由题意知，当$OG⊥OE$时，$∠GOE = 90^{\circ}$。分两种情况：当OG在OE上方时，如图①，$∠GOF = ∠GOE - ∠FOE = 90^{\circ} - 70^{\circ} = 20^{\circ}$；当OG在OE下方时，如图②，$∠GOF = ∠GOE + ∠FOE = 90^{\circ} + 70^{\circ} = 160^{\circ}$。
综上可知，$∠FOG$的度数为$20^{\circ}$或$160^{\circ}$。

2. (2025·盐城期末)已知直线$AB与CD相交于点O$,且$OM平分\angle AOC$,$\angle AOE = 90^{\circ}$。
(1)如图①,若$ON平分\angle BOC$,求$\angle MON$的度数；
(2)如图②,若$\angle CON= \frac{1}{3}\angle EON(\angle EON\lt180^{\circ})$,$\angle MON = 80^{\circ}$,求$\angle BON$的度数。

答案： 2.
(1)因为OM平分$∠AOC$，ON平分$∠BOC$，所以$∠MOC = \frac{1}{2}∠AOC$，$∠NOC = \frac{1}{2}∠BOC$，所以$∠MOC + ∠CON = \frac{1}{2}(∠AOC + ∠BOC)$，所以$∠MON = \frac{1}{2}∠AOB = \frac{1}{2}×180^{\circ} = 90^{\circ}$。
(2)设$∠BON = x^{\circ}$，因为$∠AOE = 90^{\circ}$，所以$∠BOE = 90^{\circ}$，所以$∠EON = 90^{\circ} + x^{\circ}$，所以$∠CON = \frac{1}{3}∠EON = 30^{\circ} + \frac{1}{3}x^{\circ}$。因为$∠MON = 80^{\circ}$，所以$∠COM = 80^{\circ} - (30^{\circ} + \frac{1}{3}x^{\circ}) = 50^{\circ} - \frac{1}{3}x^{\circ}$。因为OM平分$∠AOC$，所以$∠AOM = ∠COM = 50^{\circ} - \frac{1}{3}x^{\circ}$。因为$∠AOM + ∠BON = 100^{\circ}$，所以$50^{\circ} - \frac{1}{3}x^{\circ} + x^{\circ} = 100^{\circ}$，解得$x = 75$，所以$∠BON = 75^{\circ}$。

3. 已知：射线$OD在\angle AOB$内部,$OE平分\angle AOD$。
(1)如图①,试说明$\angle AOB-\angle EOB= \angle DOE$；
(2)如图②,作$OF平分\angle AOB$,试说明$\angle EOF= \frac{1}{2}\angle DOB$；
(3)如图③,在(2)的条件下,当$\angle AOD = 90^{\circ}$时,作射线$OA的反向延长线OC$,$OH在OA$的下方,且$\angle AOH= \angle AOE$,反向延长射线$OE得到射线OQ$,射线$OP在\angle HOQ$内部,$OG是\angle EOP$的平分线,若$\angle BOC-\angle DOF = 26^{\circ}$,$5\angle GOH + 2\angle POQ-\angle EOF = 71^{\circ}$,求$\angle BOP$的度数。

答案： 3.
(1)因为OE平分$∠AOD$，所以$∠AOE = ∠DOE = \frac{1}{2}∠AOD$，所以$∠AOB - ∠EOB = ∠AOE = ∠DOE$。
(2)因为OF平分$∠AOB$，所以$∠AOF = ∠BOF = \frac{1}{2}∠AOB$，所以$∠EOF = ∠AOF - ∠AOE = \frac{1}{2}∠AOB - \frac{1}{2}∠AOD = \frac{1}{2}(∠AOB - ∠AOD) = \frac{1}{2}∠DOB$。
(3)设$∠DOF = α$，$∠BOD = β$，因为OF平分$∠AOB$，所以$∠AOF = ∠BOF = α + β$。因为$∠AOD = 90^{\circ}$，所以$2α + β = 90^{\circ}$，$∠BOC = 90^{\circ} - β$。因为$∠BOC - ∠DOF = 26^{\circ}$，所以$90^{\circ} - β - α = 26^{\circ}$，所以$α = 26^{\circ}$，$β = 38^{\circ}$，即$∠DOF = 26^{\circ}$，$∠BOD = 38^{\circ}$，$∠BOC = 90^{\circ} - 38^{\circ} = 52^{\circ}$。因为$∠AOD = 90^{\circ}$，OE平分$∠AOD$，所以$∠AOE = 45^{\circ}$，$∠EOF = 19^{\circ}$。因为$∠AOH = ∠AOE$，所以$∠AOH = ∠AOE = 45^{\circ}$。因为OG是$∠EOP$的平分线，所以$∠EOG = ∠POG = 90^{\circ} - ∠GOH$。因为反向延长射线OE得到射线OQ，所以$∠EOP + ∠POQ = 180^{\circ}$，所以$2(90^{\circ} - ∠GOH) + ∠POQ = 180^{\circ}$，所以$∠POQ = 2∠GOH$。因为$5∠GOH + 2∠POQ - ∠EOF = 71^{\circ}$，所以$5∠GOH + 4∠GOH - 19^{\circ} = 71^{\circ}$，所以$∠GOH = 10^{\circ}$，所以$∠POQ = 20^{\circ}$，所以$∠BOP = 52^{\circ} + 45^{\circ} + 20^{\circ} = 117^{\circ}$。