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Given:\begin{aligned}\theta & =\frac{\pi}{6} \\\frac{d \theta}{d t} & =0.1 \mathrm{rad} / \mathrm{min} .\end{aligned}let h be the height of the air balloon above the ground. Let \theta be the angle between the ine of sight.\tan \theta=\frac{h}{3}Differentiate both side with respect to ' t '\begin{aligned}& \frac{d}{d t}(\tan \theta)=\frac{d}{d t}\left(\frac{h}{3}\right) \\\Rightarrow & \sec ^{2} \theta \frac{d \theta}{d t}=\frac{1}{3} \frac{d h}{d t} \\\Rightarrow & \frac{d h}{d t}=3 \sec ^{2} \theta \frac{d \theta}{d t}\end{aligned}\text { at } \begin{array}{l} \theta=\frac{\pi}{6}, \frac{d \theta}{d t}=0.1 \mathrm{rad} / \mathrm{min} \\\begin{aligned}\frac{d h}{d t} & =3 \sec ^{2}\left(\frac{\pi}{6}\right)(0.1) \\& =3\left(\frac{2}{\sqrt{3}}\right)^{2}\left(\frac{1}{10}\right) \\& =\frac{4}{10}=\frac{2}{5}=0.4\end{aligned}\end{array}The balloon isrising the rate of 0.4 miles/min ...