Related rates - How fast is the balloon rising? 

Let $\theta$ be the angle between the observer's line-of-sight and the horizontal, and $h$ be the height of the balloon. Then 
\[\tan \theta= \frac{h}{4}\]
\[\Rightarrow  h=4\tan \theta       (\text{in miles}).\]
Hence
\[\frac{dh}{dt}=4 (\tan \theta)  \frac{d\theta}{dt}=4 \sec^2 \theta \frac{d\theta}{dt}.\]
So we have 
\[\frac{dh}{dt}=4 \sec^2 \theta \frac{d\theta}{dt}=4 \sec^2 (\frac{\pi}{5}) \times 0.1 \]
\[=4 (1.2360)^2 \times 0.1=0.6110  \text{mile/min}.\]
So the baloon is rising with an speed about $0.61$ mile/min.