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A plane flies horizontally at an altitude of $ 5 km $ and passes directly over a tracking telescope on the ground. When the angle of elevation is $ \pi/3, $ this angle is decreasing at a rate of $ \pi/6 rad/min. $ How fast is the plane traveling at that time?

$\frac{10}{9} \pi \mathrm{km} / \mathrm{min}[\approx 130 \mathrm{mi} / \mathrm{h}]$

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Now here in the situation we are given that a plane is flying horizontally at an altitude of five kilometers per second and it passes directly over a tracking telescope on the ground. So this level we have taken us our ground now having the angle of elevation. Northern tita is by by three. This angle is decreasing at a rate of five by six radiant permanent. So they tita over DT it is decreasing decreasing means minus five by six radiant permanent. This is what we are given in the question and we have to calculate the velocity of the plane at the given instant. Now, if the right angle triangle and from this we can right that X will be equals two five divide by two engine theater. And that can also Britain us five goettingen theater. Now, if you differentiate both the function with respect to both the side with respect to T. So we'll get dX over D T will be equal to minus five Kosik and square Tito data over DT. Now we are given that data is by by three then day tater over DT we have minus by by six so you can plug this value here so we'll get dx over D D and D X over duty means the velocity, the rate of change of displacement is velocity. So that will be no minus five. Corsican square meters of Corsican square feet away. Mrs Cause I can't total, we have by by three multiplied with now dated at over data we have minus by by six. Now we can simplify this and look at the answer to be 10 by by nine kilometre permanent. So this is the velocity of the plane at the given instant. So I hope you got a problem. Thank you.