# Thread: Heights and distance problem

1. ## Heights and distance problem

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.

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

3. Hello, vivek_master146!

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!
Code:
                  A
*
* |
*   |
*     |h
*       |
* 45d     |
O * - - - - - *B
h
$\angle AO\!B \,=\,45^o,\;h \,=\,AB \,=\,OB$

Code:
                  A     2     C
. - - - - - *
. .       *   |
.   .   *       |
.     *h          |h
.   *   .           |
. * 30d   .           |
O * - - - - - * - - - - - *
h _  B     2     D
: - - - √3h - - - - - - :

$\text{In 20 seconds, it flies 2 km from }A\text{ to }C$
. . $AC\,=\,BD\,=\,22,\;\angle CO\!D \,=\,\angle 30^o,\;C\!D \,=\,h \quad\Rightarrow\quad O\!D \:=\:h + 2$

$\text{Since }\Delta CO\!D\text{ is a 30-60 right triangle, }\;O\!D \,=\,\sqrt{3}\,h.$

$\text{We have: }\:\sqrt{3}\,h \:=\:h + 2 \quad\Rightarrow\quad \sqrt{3}\,h - h \:=\:2 \quad\Rightarrow\quad (\sqrt{3}-1)h \:=\:2$

$\text{Therefore: }\;h \;=\;\frac{2}{\sqrt{3}-1}\text{ km.}$