Question

A $1.2m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is ${60}^o$. After some time, the angle of elevation reduces to ${30}^{o}C$ (See figure). Find the distance travelled by the balloon during the interval.

Answer 1

R.E.F image

In $△OCD$

$tan{60}^{\circ }=\frac{(88.2-1.2)}{x}$

(we know, $tan\theta =\frac{perpendicular}{Base})$

$tan{60}^{\circ }=\frac{87}{x}$

$=\sqrt{3}=\frac{87}{x}$

$=x=\frac{87}{\sqrt{3}}mt$

Trick to Reminder $\frac{P.B.P}{H:H:B}$

$P\to$ perpendicular

$B\to$ Base

In $△OAB$

$tan{30}^{\circ }=\frac{(88.2-1.2)}{x+y}$

$\frac{1}{\sqrt{3}}=\frac{87}{x+y}$

$x+y=87\sqrt{3}$

$y=87\sqrt{3}-x$

$y=87\sqrt{3}-\frac{87}{\sqrt{3}}mt$

During this interval fisstly balloon was at position

C that it was shifted by wind to position A

So, distance traversal by balloon is y mt.

$y=(87\sqrt{3}-\frac{87}{\sqrt{3}})mt$

$=\left(\frac{87\times 3-87}{\sqrt{3}}\right)=\frac{174}{\sqrt{3}}mt$

$=100.461mt$