Aplane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. when the angle of elevation is π/6, this angle is decreasing at a rate of π/4 rad/min. how fast is the plane traveling at that time?

The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y 
Find x:
x = y/tan(θ) 
So x = 3/tan(π/6) 
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt 
Since altitude remains the same, dy/dt = 0. Now... 
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0 
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min 

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