F.3 trigonometry (compass)－dkdqgav的部落格

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F.3 trigonometry (compass)

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發問:

In the figure, O and T represent the positions of Hong Kong and the centre of a typhoon respectively. At noon, the compass bearing of T from Hong Kong is S45E and T is 550 km from Hong Kong. If T moves steadily at 60km.h in the direction N50W, at what time will T be(a) S30E from Hong Kong(b) due south of Hong... 顯示更多 In the figure, O and T represent the positions of Hong Kong and the centre of a typhoon respectively. At noon, the compass bearing of T from Hong Kong is S45E and T is 550 km from Hong Kong. If T moves steadily at 60km.h in the direction N50W, at what time will T be (a) S30E from Hong Kong (b) due south of Hong Kong (c) nearest Hong Kong image :http://s04110118.mfs1.edu.hk/math.jpg

最佳解答:

O=HK , T=Typhoon Let P be the position of the typhoon at S30E a) Angle PTO=5 since T is S45E from O and move N50W to O with speed 60km/h. Angle POT=15 since P is S30E from O and the original position is S45E from O Angle OPT = 180 - 5 - 15 = 160 OT = 550km The distance of PO is Sin160 / 550 = sin15 / PO PO = 416.2km Time need to travel 416.2km is 416.2 / 60 = 6.937h 6.937h = 6h 56m 13s ( (6.937 -6)x60 = 56.22min.) ( (56.22-56)x60 = 13.2second) Therefore he time of the typhoon at S30E from HK is 18:56:13 (12(pm) + 6h 56m 13s) b) Let the point W at west from T and south from O & N be the location of typhoon due to the south of HK TW = 550 cos45 = 388.91km Angle WTN = 40 TN cos40 = TW TN = 388.91 / cos40 = 507.686km Time need to travel 507.686km is 507.686 / 60 = 8.461h 8.461h = 8h 27m 41s Therefore he time of the typhoon at south from HK is 20:27:41 (12(pm) + 8h 27m 41s) c) Let F be the point of the typhoon nearest HK Since the nearest point from O is perpendicular to FT, therefore angle OFT is 90 & Angle FTO is 5 & angle TOF is 85 OT cos5 = FT FT = 550 cos5 = 547.91km Time need 547.91 / 60 = 9.132h = 9h 7m 55s The time is 21h 7m 55s

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