We have that for the Question, it can be said that
From the question we are told
From,
[tex]tan\theta = \frac{h}{2}[/tex]
differentiate with respect to h
[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]
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