F.3 Trigonometry
2009-03-14 3:44 am
圖片參考：http://hk.geocities.com/poonroger12/maths.JPG
As shown in the above figure, a ballon is located at point A at noon. At that time, the balloon is fixed at a position 50m above the ground by a rope tied at point P. The angle of elevation from point P to point A is 20degree. Suddenly, the wind causes the balloon to fly up to point B. Given that X and Y are vertical projections of A and B respectively with XY = 18m.
Please find the height of B above the ground.
(The figure is not drawn in scale.)
回答 (2)
2009-03-14 5:57 am
✔ 最佳答案
50/AP = sin 20, AP = 50/sin 20 = BP...................(1)50/PX = tan 20, PX = 50/tan 20
PY = PX - 18 = 50/tan 20 - 18.................(2)
BY^2 + PY^2 = BP^2
BY^2 + (50/tan 20 - 18)^2 = (50/sin 20)^2
BY^2 + 14250.12106 = 2137.58042
BY^2 = 7121.459363
BY = 84.388 = 84.4 (3 sig. fig.)
2009-03-14 5:57 am
圖片參考：http://hk.geocities.com/poonroger12/maths.JPG
PY/PX=tan 20
=>x/[50x/(x+18)]=tan20
=>x+18=50 tan20
=>x=50 tan 20-18--(1)
AP=BP
=>AP=[(50 tan 20)^2+50^2]^(1/2)=50 sec 20= 122.5 m//
2009-03-13 22:00:23 補充：
=>BP=[(50 tan 20)^2+50^2]^(1/2)=50 sec 20
BP^2=PY^2+YB^2
=>(50 sec 20)^2=(50 tan 20-19)^2+YB^2
=>YB=80 m//
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090313000051KK01402