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

(N/A) Let the initial position of the balloon be $A$ and its final position be $B$. Let $CD$ be the girl,where $CD = 1.2\, m$.
The height of the balloon from the ground is $88.2\, m$. The height of the balloon from the eye level of the girl is $88.2\, m - 1.2\, m = 87\, m$.
Let the eye level of the girl be the horizontal line $CG$. Thus,$AE = BG = 87\, m$.
In $\triangle ACE$,$\tan 60^{\circ} = \frac{AE}{CE} \implies \sqrt{3} = \frac{87}{CE} \implies CE = \frac{87}{\sqrt{3}} = 29\sqrt{3}\, m$.
In $\triangle BCG$,$\tan 30^{\circ} = \frac{BG}{CG} \implies \frac{1}{\sqrt{3}} = \frac{87}{CG} \implies CG = 87\sqrt{3}\, m$.
The distance travelled by the balloon is $EG = CG - CE = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\, m$.