From a balloon vertically above a straight road,the angles of depression of two cars at an instant are found to be $45^{\circ}$ and $60^{\circ}$. If the cars are $100 \, m$ apart,find the height of the balloon.

(N/A) Let the height of the balloon at $P$ be $h$ meters. Let $A$ and $B$ be the two cars. Thus $AB = 100 \, m$. Let $Q$ be the point on the road directly below the balloon $P$.
In $\triangle PAQ$,$\angle PAQ = 45^{\circ}$. Therefore,$\tan 45^{\circ} = \frac{PQ}{AQ} \implies 1 = \frac{h}{AQ} \implies AQ = h$.
In $\triangle PBQ$,$\angle PBQ = 60^{\circ}$. Therefore,$\tan 60^{\circ} = \frac{PQ}{BQ} \implies \sqrt{3} = \frac{h}{BQ} \implies BQ = \frac{h}{\sqrt{3}}$.
Since $AQ = AB + BQ$,we have $h = 100 + \frac{h}{\sqrt{3}}$.
$h - \frac{h}{\sqrt{3}} = 100 \implies h \left( \frac{\sqrt{3}-1}{\sqrt{3}} \right) = 100$.
$h = \frac{100 \sqrt{3}}{\sqrt{3}-1} = \frac{100 \sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{100(3+\sqrt{3})}{2} = 50(3+\sqrt{3}) \, m$.
Thus,the height of the balloon is $50(3+\sqrt{3}) \, m$ (approximately $236.6 \, m$).