Answer:
-0.584 rad/hr, rounded to the nearest 1000th
Explanation:
Let D = D(t) = the horizontal distance from the observer to the plane at time t in units of km.
Let E = E(t) = the angle of elevation at time t in units of Radians.
Then, by geometry,
tan(E) = 16 / D, and furthermore, we have that
tan(E) = 16 * D^(-1)
d/dt (tan(E)) = d/dt ( 16 * D^(-1))
sec^2 (E) * dE/dt = 16(-1)(D^(-2)) * dD/dt
( sqrt(D^2 + 16^2) / D )^2 * dE/dt = -16 (D^(-2)) * dD/dt
(D^2 + 16^2)/D^2 * dE/dt = -16 (D^(-2)) * dD/dt
dE/dt = -16 (D^(-2)) * dD/dt * D^2 / (D^2 + 16^2)
dE/dt = -16 (dD/dt) / (D^2 + 16^2)
dE/dt = -16 (580 km/hr) / (125^2 + 16^2), at time t such that D(t) = 125 km,
dE/dt = (-9280 / 15881) rad/hr
dE/dt = -0.584 rad/hr, rounded to the nearest 1000th