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a)i) FF’ = EFsin30° = 20(0.5) = 10 m AF’ = FF’ / tan60° = 10 / √3 EF’ = EFcos30° = 10√3 AE2 = AF’ 2 + EF’ 2 = 100 / 3 + 300 = 1000 / 3 ∴ AE = √(1000 / 3) = (10√30) / (3) ii) AF = 10 / sin60° = 20 / √3 Consider △AEF, cos∠AEF = (AE2 + EF’ 2 - AF’ 2) / 2(AE)(EF) = (1000 / 3 + 400 – 400 / 3) / 2[(10√30) / (3)](20) = 600 / (400√30) / (3) ∴ ∠AEF = 34.75634… = 34.8° (3 sig. fig.) b) Consider △BEF, ∠FEB = 180° - ∠AEF (adj. ∠ on st. line) = 180° - 34.75634° = 145.24366° And ∠FBE = 180° - ∠FEB - ∠EFB (∠s sum of △) = 180° - 145.24366° - 20° = 14.75634° By sine law, FB / sin∠FEB = EF / sin∠FBE FB / sin145.24366° = 20 / sin14.75634° ∴ FB = 44.76385 m And EB / sin∠EFB = EF / sin∠FBE EB / sin20° = 20 / sin14.75634° ∴ EB = 26.85576 m ∴Time taken for the red car to travel from F to B = 44.76385 / 3 = 14.92128 = 14.9 s (3 sig. fig.) And Time taken for the yellow car to travel from E to B = 26.85576 / 2 = 13.42788 = 13.4 s (3 sig. fig.) ∴ The yellow car will reach B before the red one. 2009-04-15 18:34:09 補充: Sorry, 我最未搞亂左 ∴Time taken for the yellow car to travel from F to B = 44.76385 / 3 = 14.92128 = 14.9 s (3 sig. fig.) And Time taken for the red car to travel from E to B = 26.85576 / 2 = 13.42788 = 13.4 s (3 sig. fig.) ∴ The yellow car will not reach B before the red one. 2009-04-16 09:36:07 補充: AE2 = AF’ 2 + EF’ 2 should be AE^2 = AF'^ 2 + EF'^2 2009-04-16 09:36:52 補充: cos∠AEF = (AE2 + EF’ 2 - AF’ 2) / 2(AE)(EF) should be cos∠AEF = (AE^2 + EF'^ 2 - AF'^2) / 2(AE)(EF)

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