5. (2025·盐城期末)【问题情境】阅读资料:光遇到水面、玻璃以及其他许多物体的表面都会发生反射.如图①,经过入射点O且垂直于反射面的直线OE叫作法线.入射光线CO与法线的夹角∠COE叫作入射角.反射光线OD与法线的夹角∠EOD叫作反射角.
光的反射定律:在反射现象中,反射光线、入射光线和法线都在同一个平面内:反射光线、入射光线分居法线两侧;反射角等于入射角,即∠EOD=∠COE.
(1)探究图①中入射光线与镜面所夹的锐角∠AOC与反射光线与镜面所夹的锐角∠BOD的数量关系,并说明理由.
【结论应用】请用【问题情境】中获得的结论解决以下问题:
如图②,直线MN//PQ,点A在直线PQ上,点C在直线MN上,光线AC被MN反射后再次被PQ反射,入射光线AC经过两次反射的光线为BD,其中点B在直线PQ上.
(2)BD与AC有怎样的位置关系?并说明理由.
(3)如图③,已知∠1=66°,直线MN绕点C顺时针旋转α(0°＜α＜30°)至直线GH,当α为何值时,BD//GH.
(4)直线MN绕点C顺时针旋转α(0°＜α＜180°),直线BD与直线AC相交于点E,请直接写出∠AEB和α之间的数量关系.
光的反射定律:在反射现象中,反射光线、入射光线和法线都在同一个平面内:反射光线、入射光线分居法线两侧;反射角等于入射角,即∠EOD=∠COE.
(1)探究图①中入射光线与镜面所夹的锐角∠AOC与反射光线与镜面所夹的锐角∠BOD的数量关系,并说明理由.
【结论应用】请用【问题情境】中获得的结论解决以下问题:
如图②,直线MN//PQ,点A在直线PQ上,点C在直线MN上,光线AC被MN反射后再次被PQ反射,入射光线AC经过两次反射的光线为BD,其中点B在直线PQ上.
(2)BD与AC有怎样的位置关系?并说明理由.
(3)如图③,已知∠1=66°,直线MN绕点C顺时针旋转α(0°＜α＜30°)至直线GH,当α为何值时,BD//GH.
(4)直线MN绕点C顺时针旋转α(0°＜α＜180°),直线BD与直线AC相交于点E,请直接写出∠AEB和α之间的数量关系.
答案：
(1)∠AOC = ∠BOD,理由如下:
∵OE⊥AB,
∴∠AOE = ∠BOE = 90°.
∵∠COE = ∠DOE,
∴∠AOE - ∠COE = ∠BOE - ∠DOE,即∠AOC = ∠BOD.
(2)BD//AC,理由如下:
∵MN//PQ,
∴∠BCM = ∠CBQ,由(1)的结论可得∠ACN = ∠BCM,∠CBQ = ∠PBD,
∴∠ACB = 180° - ∠ACN - ∠BCM = 180° - 2∠BCM,∠CBD = 180° - ∠CBQ - ∠PBD = 180° - 2∠CBQ,
∴∠ACB = ∠CBD,
∴AC//BD.
(3)
∵MN//PQ,
∴∠NCA = ∠1 = 66°.
∵直线MN绕点C顺时针旋转α(0°＜α＜30°)至直线GH,即∠HCN = α,
∴∠ACH = ∠ACN - ∠HCN = 66° - α,由(1)的结论可得∠BCG = ∠ACH = 66° - α,
∴∠ACB = 180° - ∠ACH - ∠BCG = 180° - (66° - α) - (66° - α) = 48° + 2α,
∴∠ABC = 180° - ∠1 - ∠ACB = 180° - 66° - (48° + 2α) = 66° - 2α,由(1)的结论可得∠DBP = ∠ABC = 66° - 2α,
∴∠CBD = 180° - ∠DBP - ∠ABC = 180° - (66° - 2α) - (66° - 2α) = 48° + 4α.
∵当∠BCG + ∠CBD = 180°,即66° - α + 48° + 4α = 180°时,BD//GH,
∴α = 22°.
(4)∠AEB = 2α或∠AEB = 360° - 2α.解析:分两种情况讨论:
①如图①,当BD在AC的左侧时,直线BD与AC相交于点E.
∵MN//PQ,
∴∠NCA = ∠1.
∵直线MN绕点C顺时针旋转α,即∠NCH = ∠MCG = α,
∴∠2 = ∠NCA - ∠NCH = ∠1 - α,由(1)的结论可得∠GCB = ∠2 = ∠1 - α,
∴∠3 = ∠GCB - ∠GCM = ∠1 - 2α.
∵MN//PQ,
∴∠4 = ∠3 = ∠1 - 2α,由(1)结论有∠5 = ∠4 = ∠1 - 2α,
∴∠6 = ∠5 = ∠1 - 2α,
∴∠E = ∠1 - ∠6 = ∠1 - (∠1 - 2α) = 2α.
②如图②,当BD在AC的右侧时,直线BD与AC相交于点E.
∵MN//PQ,
∴∠MCA = 180° - ∠1.
∵直线MN绕点C顺时针旋转α,即∠MCG = α,
∴∠2 = 180° - ∠MCG = 180° - α,
∴∠3 = ∠MCA - ∠2 = 180° - ∠1 - (180° - α) = α - ∠1,由(1)结论可得,∠GCB = ∠3 = α - ∠1.又∠4 = ∠2 = 180° - α,
∴∠5 = ∠GCB - ∠4 = α - ∠1 - (180° - α) = 2α - ∠1 - 180°.
∵MN//PQ,
∴∠6 = ∠5 = 2α - ∠1 - 180°,由(1)结论可得∠7 = ∠6 = 2α - ∠1 - 180°,
∴∠ABE = ∠7 = 2α - ∠1 - 180°.
∵∠BAE = ∠1,
∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - (2α - ∠1 - 180°) - ∠1 = 360° - 2α.
综上所述,∠AEB = 2α或∠AEB = 360° - 2α.
(1)∠AOC = ∠BOD,理由如下:
∵OE⊥AB,
∴∠AOE = ∠BOE = 90°.
∵∠COE = ∠DOE,
∴∠AOE - ∠COE = ∠BOE - ∠DOE,即∠AOC = ∠BOD.
(2)BD//AC,理由如下:
∵MN//PQ,
∴∠BCM = ∠CBQ,由(1)的结论可得∠ACN = ∠BCM,∠CBQ = ∠PBD,
∴∠ACB = 180° - ∠ACN - ∠BCM = 180° - 2∠BCM,∠CBD = 180° - ∠CBQ - ∠PBD = 180° - 2∠CBQ,
∴∠ACB = ∠CBD,
∴AC//BD.
(3)
∵MN//PQ,
∴∠NCA = ∠1 = 66°.
∵直线MN绕点C顺时针旋转α(0°＜α＜30°)至直线GH,即∠HCN = α,
∴∠ACH = ∠ACN - ∠HCN = 66° - α,由(1)的结论可得∠BCG = ∠ACH = 66° - α,
∴∠ACB = 180° - ∠ACH - ∠BCG = 180° - (66° - α) - (66° - α) = 48° + 2α,
∴∠ABC = 180° - ∠1 - ∠ACB = 180° - 66° - (48° + 2α) = 66° - 2α,由(1)的结论可得∠DBP = ∠ABC = 66° - 2α,
∴∠CBD = 180° - ∠DBP - ∠ABC = 180° - (66° - 2α) - (66° - 2α) = 48° + 4α.
∵当∠BCG + ∠CBD = 180°,即66° - α + 48° + 4α = 180°时,BD//GH,
∴α = 22°.
(4)∠AEB = 2α或∠AEB = 360° - 2α.解析:分两种情况讨论:
①如图①,当BD在AC的左侧时,直线BD与AC相交于点E.
∵MN//PQ,
∴∠NCA = ∠1.
∵直线MN绕点C顺时针旋转α,即∠NCH = ∠MCG = α,
∴∠2 = ∠NCA - ∠NCH = ∠1 - α,由(1)的结论可得∠GCB = ∠2 = ∠1 - α,
∴∠3 = ∠GCB - ∠GCM = ∠1 - 2α.
∵MN//PQ,
∴∠4 = ∠3 = ∠1 - 2α,由(1)结论有∠5 = ∠4 = ∠1 - 2α,
∴∠6 = ∠5 = ∠1 - 2α,
∴∠E = ∠1 - ∠6 = ∠1 - (∠1 - 2α) = 2α.
②如图②,当BD在AC的右侧时,直线BD与AC相交于点E.
∵MN//PQ,
∴∠MCA = 180° - ∠1.
∵直线MN绕点C顺时针旋转α,即∠MCG = α,
∴∠2 = 180° - ∠MCG = 180° - α,
∴∠3 = ∠MCA - ∠2 = 180° - ∠1 - (180° - α) = α - ∠1,由(1)结论可得,∠GCB = ∠3 = α - ∠1.又∠4 = ∠2 = 180° - α,
∴∠5 = ∠GCB - ∠4 = α - ∠1 - (180° - α) = 2α - ∠1 - 180°.
∵MN//PQ,
∴∠6 = ∠5 = 2α - ∠1 - 180°,由(1)结论可得∠7 = ∠6 = 2α - ∠1 - 180°,
∴∠ABE = ∠7 = 2α - ∠1 - 180°.
∵∠BAE = ∠1,
∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - (2α - ∠1 - 180°) - ∠1 = 360° - 2α.
综上所述,∠AEB = 2α或∠AEB = 360° - 2α.