Q:

How to find the Rate of Change?

Could anyone explain to me the process of this problem? My book tells me to find an related equation and then take the derivative of that in respect to time.

Update—Thanks for your help Mark!
So the correct method was

$tan(\theta) = \frac yx$
$\frac{d\theta}{dt}$$[tan(\theta) = \frac yx]$ = $sec^2(\theta)\frac{d\theta}{dt}$ = $\frac{\frac{dy}{dt}x - \frac{dx}{dt}y}{x^2}$
Substitute in $-700$ (distance between viewer and airplane decreases with time) for  $\frac{dx}{dt}$, $4$ for $y$, $0$ for $\frac{dy}{dt}$.
The value of $x$ is obtainable by plugging in $45°, 60°, 75°$ for $tan(\theta)$ = $\frac yx$.
The new equation is now $sec^2(\theta)\frac{d\theta}{dt}$ = $\frac{0x - (-700)4}{x^2}$ or simply $sec^2(\theta)\frac{d\theta}{dt}$ = $\frac{2800}{x^2}$.
$\frac {d\theta}{dt} = \frac{\frac{2800}{x^2}}{sec^2(\theta)}$ = $\frac{2800}{x^2sec^2(\theta)}$

A:

You're dealing with a right triangle. In this case, we are given the height (call it $y$), and we know that $y = 4$ miles. We also know the rate at which the base of the triangle (call it $x$) is changing with respect to time, namely $\frac{dx}{dt} = -700$ (it's negative because it's causing the base of the triangle to become smaller). You can relate the angle $\theta$ to $x$ and $y$ using trigonometric functions. You will have an equation that relates $\theta$ to the height $y$ and the base $x$. Then take the derivative of both sides with respect to time, plug in the values you are given, and solve. Note: you will be solving for $\frac{d\theta}{dt}$, the rate at which the angle $\theta$ is changing. Let me know if you need more help.

