Answer
$192m$
Work Step by Step
Step 1 Draw a diagram as shown in the figure, where T is the position of the balloon. and we know three angles and a distance (PQ=60m) as shown.
Step 2. In the right triangle of TQS, $\angle TQS=71^\circ$ so we have $\angle QTS=90-71=19^\circ$
Step 3. In the right triangle of TPR, $\angle TPR=62^\circ$ so we have $\angle PTR=90-62=28^\circ$ and
$\angle PTQ=28-19=9^\circ$
Step 4, In triangle PTQ, $\angle TPQ=62-32=30^\circ$ use the Law of Sine, we have $\frac{QT}{sin30^\circ}=\frac{60}{sin9^\circ}$ thus the distance from Q to the balloon $QT=60\times\frac{sin30^\circ}{sin9^\circ}\approx192m$
