Figure 10 shows a circle with centre O and radius 10 m on a vertical
wall which stands on the horizontal ground. A , B and C are three
points on the circumference of the circle such that A is vertically below
O , ∠AOB = 90° and ∠AOC = 20° .
A laser emitter D on the ground shoots a laser beam at B . The laser
beam then sweeps through an angle of 30° to shoot at A . The angles of
elevation of B and A from D are 60° and 30° respectively.
(a) Let A be h m above the ground.
(i) Express AD and BD in terms of h .
(ii) Find h .
(b) Another laser emitter E on the ground shoots a laser beam at A with angle of elevation 25° . The laser beam then sweeps through an angle of 5° to shoot
at C . Find ∠ACE .
有圖:
http://photie.com/photos/original/10387.jpg
成題:
http://photie.com/photos/medium/10386.jpg
hkcee maths I 2000#17
2008-07-29 4:34 am
回答 (2)
2008-07-30 1:09 am
✔ 最佳答案
ai) AD = h/sin30o = 2h
BD = h/sin60o = 2(h+10)/sqrt(3)
aii)
AB^2 = 100 + 100 (Pyth. theorem)
AB^2 = 200
Also ,
AB^2 = AD^2 + BD^2 - 2(AD)(BD)cos30o - cosine formula
200 = (2h)^2 + [2(h+10)/sqrt(3)]^2 - 2[2h][2(h+10)/sqrt(3)](cos30o)
通分母
200 = { (3)(2h)^2 + [2(h+10)]^2 - (3)(2)(2h)[2(h+10)/sqrt(3)][sqrt(3)/2] } /3
600 = 12h^2 + 4(h^2 + 20h + 100) - 12h^2 - 120h
600 = 4h^2 - 40h + 400
4h^2 - 40h - 200 = 0
h^2 - 10h - 50 = 0
h= 13.66(m) / -3.66 (rejected)
b)
AC = sqrt[100 + 100 - 2(10)(10)(cos20o)]
AC = sqrt(200)sqrt(1-cos20o)
AE = h/sin25o
By the sine formula,
AE / sin∠ACE = AC / sin5o
(h/sin25o)(sin5o) = AC (sin∠ACE )
sin∠ACE = 0.8111598
∠ACE = 54.21o
2008-10-09 8:18 am
收錄日期: 2021-04-11 00:44:27
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