A plane flies horizontally at an altitude 5 km and passes directly over a tracki

Question

A plane flies horizontally at an altitude 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 radians per minute. How fast is the plane traveling at that time? Which equation below is appropriate to use to solve this problem?(Circle your answer) x^2 + y^2 = z^2 x^2 + 5^2 = z^2 tan(theta) = 5/pi tan (theta) = x/5 Solve the problem. For full credit, you must show work that clearly communicates how you arrived at your answer and uses proper mathematical notation. Write the answer in a sentence to indicate its meaning in the context of the problem.

Explanation / Answer

Let x = horizontal distance of the plane from the telescope when the angle of elevation is
=> 5/x =tan

==> x/5 = cot

==>
=> (1/5) dx/dt = -cosec^2 * d/dt
=> horizontal speed at which the plane is flying is
dx/dt
= -5cosec^2 * (rate of change of angle of elevation)
= -5 cosec^2 (/3) * (-/6)
= 10/6 km/min.

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