#### Answer

The balloon is $0.485mi$ above the ground.

#### Work Step by Step

Take a look at the image below.
We have $a\parallel AB$. Now recall the characteristics of pairs of angles in parallel lines. As pairs of alternate internal angles are equal,
$$\angle A=35^\circ\hspace{2cm}\angle B=31^\circ$$
Also, as $a\parallel AB$, $$\angle L +35^\circ+31^\circ=180^\circ$$
$$\angle L+66^\circ=180^\circ$$
$$\angle L=114^\circ$$
$$\angle A=35^\circ\hspace{.75cm}\angle B=31^\circ\hspace{.7cm}\angle L=114^\circ\hspace{.75cm}AB=1.5mi\hspace{.75cm}LH=?$$
1) Analysis
- To find $LH$, we first need to find $LB$.
- To find $LB$, we would use law of sines. 2 groups of side and its opposite angles must be known.
- The opposite angles of $LB$ and $AB$, $\angle A$ and $\angle L$ respectively, are already known.
2) Apply law of sines to find $LB$.
- The opposite angle of $LB$ is $\angle A$, $\sin A=\sin 35^\circ\approx0.57$
- $AB=1.5mi$, its opposite angle is $\angle L$, $\sin L=\sin 114^\circ\approx0.91$
According to law of sines:
$$\frac{LB}{\sin A}=\frac{AB}{\sin L}$$
$$LB=\frac{AB\sin A}{\sin L}$$
$$LB=\frac{1.5mi\times0.57}{0.91}$$
$$LB\approx0.94mi$$
4) Find $LH$
In $\triangle LHB$: $LH \bot HB$
$$LH=LB\sin B$$
$$LH=0.94mi\times\sin 31^\circ$$
$$LH\approx0.485mi$$
Therefore, the balloon is $0.485mi$ above the ground.