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標題:
5條5分是~~~~~~~~~~~~~~
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最佳解答:
(1) Angle ACB = Angle ADB = 55 ( Angles in same segment ) Angle ABC = Angle ACB = 55 ( Base angles, isos. triangle ) Angle BAC = 180 - Angle ABC - Angle ACB = 70 Angle BDC = Angle BAC = 70 ( Angles in same segment ) (2) Angle BCD = Angle DAE = 100 ( ext. angle, cyclic quad. ) Angle CBD = 180 - Angle BCD - Angle BDC = 35 Arc DA = 2 x Arc CD Angle ABD = 2 x Angle CBD ( arcs prop. to angles at circumference ) Angle ABD = 2 x 35 = 70 (3) Angle ABC = Angle ADC = 30 ( Angles in same segment ) Angle ACB = 90 ( Angle in semicircle ) Angle BAC = 180 - Angle ACB - Angle ABC = 60 Arc AB : Arc BC : Arc CA = Angle ACB : Angle BAC : Angle ABC ( Arcs prop. to angles at circumference ) = 3 : 2 : 1 (4a) Angle ABE = Angle ACE + Angle BEC ( ext. angle of triangle ) Angle BEC = 20 Angle BAD = Angle BEC = 20 ( Angles in the same segment ) (4b) Angle AFB = 180 - Angle BAD - Angle ABE = 105 Angle DFE = Angle AFB = 105 (4c) Angle EMF = 90 ( line from centre to mid-pt. of chord perp. to chord ) Angle MEF + Angle EMF = Angle DFE ( ext. angle of triangle ) Angle MEF = 15 (5a) Angle ODE = Angle OED = 70 ( base angles, isos. triangle ) Angle DOE = 180 - Angle ODE - Angle OED = 40 Arc CD : Arc DE = 3 : 2 Angle COD : Angle DOE = 3 : 2 ( arcs prop. to angles at centre ) Angle COD = 60 (5b) Angle BOC = Angle DOE = 40 ( equal chord, equal angles ) Angle BOE = Angle BOC + Angle COD + Angle DOE
其他解答:
5條5分是~~~~~~~~~~~~~~
發問:
此文章來自奇摩知識+如有不便請留言告知
[IMG]http://i293.photobucket.com/albums/mm67/zaza520/0001-65.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0003-47.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0006-35.jpg[/IMG]最佳解答:
(1) Angle ACB = Angle ADB = 55 ( Angles in same segment ) Angle ABC = Angle ACB = 55 ( Base angles, isos. triangle ) Angle BAC = 180 - Angle ABC - Angle ACB = 70 Angle BDC = Angle BAC = 70 ( Angles in same segment ) (2) Angle BCD = Angle DAE = 100 ( ext. angle, cyclic quad. ) Angle CBD = 180 - Angle BCD - Angle BDC = 35 Arc DA = 2 x Arc CD Angle ABD = 2 x Angle CBD ( arcs prop. to angles at circumference ) Angle ABD = 2 x 35 = 70 (3) Angle ABC = Angle ADC = 30 ( Angles in same segment ) Angle ACB = 90 ( Angle in semicircle ) Angle BAC = 180 - Angle ACB - Angle ABC = 60 Arc AB : Arc BC : Arc CA = Angle ACB : Angle BAC : Angle ABC ( Arcs prop. to angles at circumference ) = 3 : 2 : 1 (4a) Angle ABE = Angle ACE + Angle BEC ( ext. angle of triangle ) Angle BEC = 20 Angle BAD = Angle BEC = 20 ( Angles in the same segment ) (4b) Angle AFB = 180 - Angle BAD - Angle ABE = 105 Angle DFE = Angle AFB = 105 (4c) Angle EMF = 90 ( line from centre to mid-pt. of chord perp. to chord ) Angle MEF + Angle EMF = Angle DFE ( ext. angle of triangle ) Angle MEF = 15 (5a) Angle ODE = Angle OED = 70 ( base angles, isos. triangle ) Angle DOE = 180 - Angle ODE - Angle OED = 40 Arc CD : Arc DE = 3 : 2 Angle COD : Angle DOE = 3 : 2 ( arcs prop. to angles at centre ) Angle COD = 60 (5b) Angle BOC = Angle DOE = 40 ( equal chord, equal angles ) Angle BOE = Angle BOC + Angle COD + Angle DOE
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