Rate of change of angle of elevation

Set up the right triangle: altitude $5000$ ft is fixed, $x$ is the horizontal distance from the observer to the point below the plane, and $\theta$ is the angle of elevation. Then $\tan\theta = \dfrac{5000}{x}$. Since the plane flies toward the observer, $x$ decreases, so $\frac{dx}{dt} = -300$ mph. Converting to ft/s: $300 \text{ mph} = 300 \times \dfrac{5280}{3600} = 440$ ft/s, so $\frac{dx}{dt} = -440$ ft/s.

Differentiate $\tan\theta = \dfrac{5000}{x}$ implicitly with respect to $t$: $\sec^2\theta \cdot \frac{d\theta}{dt} = -\frac{5000}{x^2}\frac{dx}{dt}.$ At $x = 4000$, we have $\tan\theta = \dfrac{5000}{4000} = \dfrac{5}{4}$, so $\tan^2\theta = \dfrac{25}{16}$. Therefore $\sec^2\theta = 1 + \tan^2\theta = 1 + \dfrac{25}{16} = \dfrac{41}{16}$.

Substitute into the equation: $\frac{41}{16}\frac{d\theta}{dt} = -\frac{5000}{16{,}000{,}000}(-440) = \frac{5000 \times 440}{16{,}000{,}000} = \frac{2{,}200{,}000}{16{,}000{,}000} = \frac{11}{80}.$ Solving, $\frac{d\theta}{dt} = \dfrac{16}{41} \times \dfrac{11}{80} = \dfrac{176}{3280} = \dfrac{11}{205} \approx 0.0537$ rad/s.