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The angle of elevation from a point(on ground) is 45 degree and after 20 seconds its 30 degree If the aircraft is flying at a constant speed at 360 km/h. Find the height of plane from the ground.
I tried to solve the problem. The answer is 2000\sqrt{3}. Please help.
Follow the link http://img684.imageshack.us/img684/8819/unledaj.png
angle AOB=45, angle COD=30. AB=CD, CA=DB. A is the initial position of the plane; and C is the position of the plane after 20 seconds. In 20 seconds, the plane covers a distance of 2000 mitres. So AC=2000.
tan(AOB)=tan(45)=1=AB/BO. So AB=BO
tan(COD)=tan(30)=1/sqrt(3)=CD/DO=CD/(DB+BO)=CD/(2000+BO)=AB/(2000+BO)=AB/(2000+AB)
That is AB/(2000+AB) = 1/sqrt(3)
Solving, AB = 2000/{sqrt(3)-1} which is the required height.
EDIT: In case the above link is non-accessible, here is another link http://oi56.tinypic.com/2eldz4n.jpg
Hello, vivek_master146!
Quote:
The angle of elevation of a plane from a point on ground is 45 degree
and after 20 seconds it's 30 degrees.
If the plane is flying at a constant speed at 360 km/h, find the altitude of plane .
The answer is 2000√3. . Wrong!
http://latex.codecogs.com/png.latex?...,=\,AB \,=\,OBCode:A
*
* |
* |
* |h
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* 45d |
O * - - - - - *B
h
Code:A 2 C
. - - - - - *
. . * |
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. *h |h
. * . |
. * 30d . |
O * - - - - - * - - - - - *
h _ B 2 D
: - - - √3h - - - - - - :
http://latex.codecogs.com/png.latex?...}A\text{ to }C
. . http://latex.codecogs.com/png.latex?...\!D \:=\:h + 2
http://latex.codecogs.com/png.latex?...\,\sqrt{3}\,h.
http://latex.codecogs.com/png.latex?...{3}-1)h \:=\:2
http://latex.codecogs.com/png.latex?...-1}\text{ km.}