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Let \(h\) be the height of the balloon and \(\theta\) the angle the balloon makes with the horizon. |dw:1476293797819:dw| You have \(\tan\theta=\dfrac{h}{4}\). Differentiate both sides with respect to time, so that \[\sec^2\theta\frac{\mathrm d\theta}{\mathrm dt}=\frac{1}{4}\frac{\mathrm dh}{\mathrm dt}\]You're given that the angle is increasing at a rate of \(0.2\text{ rad/min}\), so \(\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{1}{5}=0.2\), and you want to find the rate at which the height is changing, or \(\dfrac{\mathrm dh}{\mathrm dt}\). You're also given a particular angle of \(\theta=\dfrac{\pi}{5}\), so you have everything you need to solve for \(\dfrac{\mathrm dh}{\mathrm dt}\).
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