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 π/3, this angle is decreasing at a rate of π/3 rad/min. How fast is the plane traveling at that time?

Answer 1

I assume you made a diagram.
I let the angle of elevation be Ø, and the horizontal distance of the plane as x km
In mine,
tanØ = x/5
xtanØ = 5
dx/dt tanØ + x sec^2 Ø dØ/dt = 0

when Ø = π/3 , (I recognize the 30-60-90° triangle)
so
cosØ = 1/2
secØ = 2
sec^2 Ø = 4

dx/dt (√3) + (5/√3)(4)(-π/3) = 0
√3 dx/dt = -(5/√3)(-π/3) = 5π/(3√3)
dx/dt = 5π/9 km/min
or
(5π/9)(60) km/h which is appr 104.7 km/h

better check my arithmetic, I did not write it on paper first.

Answer 2

From triangle,cotu=x/5 differentiate both side cosec^2udu/dt=dx/dt/5subtitute u and du/dt and solve for dx/dt