有關三角問題(數學)

[匿名]

2008-06-23 5:58 am

A,B and C are three observation points on a horizontal line . TD is a vertical tower, with the bottom D on teh same horizontal level as A, B and C. the angles of elevation of T from A, B and C are 30deg, 45deg and 60deg respectively. AB = 60m and BC = 50m. find the height of the tower

更新1:

the height of the tower is 74.16m 要過程

回答 (1)

Audrey Hepburn

2008-06-23 7:30 am

✔ 最佳答案

圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_hepburn2008/A_Hepburn01Jun222318.jpg?t=1214147975

From the figure, TD is the height of the tower. Let TD = h

AD = h / tan30* = h√3, BD = h, CD = h / tan60* = h / √3

Now, consider triangle ABD

Apply cosine law,

cosABD = (AB2 + BD2 - AD2) / 2AB X BD

cosABD = (3600 + h2 - 3h2) / 120h = (1800 - h2) / 60h ─── (1)

Consider triangle BCD,

Apply cosine law,

cosCBD = (BC2 + BD2 - CD2) / 2BC X BD

cosCBD = (2500 + h2 - h2/3) / 100h

cosCBD = (3750 + h2) / 150h ─── (2)

Since angle ABD + angle CBD = 180*

So, (2): cosCBD = cos(180* - ABD) = -cosABD

Therefore, (1800 - h2) / 60h = -(3750 + h2) / 150h

9000 - 5h2 = -7500 - 2h2

h2 = 5500

h = 74.16 (cor. to 2 d.p.) or -74.16 (rejected)

Therefore, the height of the tower is 74.16 m.

參考: Myself~~~

👍 0 👎 0