Question

A 1.2m tall girl spots a ballon moving with the wind in a horizontal line at a height of 88.2m from the ground. The angle of elevation of the ballon from the eyes of the girl at any instant is 60°. After sometime the angle of elevation reduced to 30°. Find the distance travelled by ballon during the interval.

Answer 1

put values of AD and AB and find answer

Answer 2

$ap = 1.2m \\ height \: of \: ballon = 88.2m \\ distance \: o \: ballon = ....... \\ ln \: \: triangle \: abc \\ \tan(60) = \frac{bc}{ab} \\ \sqrt{3} = \frac{87}{ab} \ \sqrt{3} ab \: = 87 \\ ab = \frac{87}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{87 \: \sqrt{3} }{3} \\ ab = 29\sqrt{3} m \\ \\ ln \: triangle \: ade \\ \tan(30) = \frac{de}{ad} \\ \frac{1}{ \sqrt{3} } = \frac{87}{ad} \\ ad = 87 \sqrt{3m} \\ (distance) \: ce = b = ad - ab \\ = 87 \sqrt{ 3} - 29 \sqrt{3} \\ = 58 \sqrt{3} m \\ 58(1.73) \\ 100.34m$
Hope this helps.

@ Neer Tanwar.

Thank you