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6條maths 6條maths6條maths

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6條maths [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0008.jpg[/IMG]

最佳解答:

(1a) ∠BDC = 30° ∠ADC = 90° ∠ADB = ∠ADC - ∠BDC = 60° (b) ∠ABC = 90° ∠ACB = 60° Arc AB : arc BC = ∠ACB : ∠BAC = 2 : 1 (c) AB = AC cos 30° BC = AC sin 30° AB : BC = cos 30° : sin 30° = √3 : 1 (2a) ∠EAB = ∠EDC and ∠EBA = ∠ECD (Angles in the same segment) So △EBA is similar to △ECD (AAA) (b) From the result of (a), we have: EB/EC = EA/ED 6/3 = 4/y y = 2 (3) ∠CDB = ∠BDA (Equal chord, equal angle at circumference) ∠BOA = 2∠BDA (Angle at centre = Twice angle at circumference) ∠CDO = ∠CDB + ∠BDA = 2∠BDA So, BO//CD for the reason of corresponding angle. (4) ∠ADC = 90° (Angle in semi-circle) ∠ADB = ∠ACB = 50° (Angle in the same segment) y = 40 x = 180 - 90 - 20 = 70 (5) ∠ACB = ∠DAC (Alt. angles AD//BC) x = 25 ∠ADB = ∠ACB = 25° (Angle in the same segment) y = 180 - 56 - 25 - 25 = 74 (6) ∠DCA = ∠DBA = 30° (Angle in the same segment) ∠ADC = 90° (Angle in semi-circle) ∠DAC = 180° - 90° - 30° = 60°

其他解答:6A560DB6A1464DEC