Related rates problem, can't find θ 
A police officer sits on the side of the road. A car passes 20 feet from her, traveling at the rate of 75 mph (110 ft/sec). How fast is
  the angle $\theta$ changing at that time?


So far, I have the base equation for $\theta$ as $\tan\theta=D/20$, where $D$ represents the distance between the 20' line and the car. Finding the derivative, and using 110 as $\frac{dD}{dt}$ gives me
$$\frac{d\theta}{dt}=\frac{110}{20\sec^2\theta}$$
The only problem is I can't seem to find a way to solve for $\theta$.
 A: If the question is just how fast theta is changing then you don't need to solve for theta, you need to solve for $$ d\theta/dt $$ which it looks like you already did.  To solve for theta explicitly here would require integration ("anti-differentiation") which you may not have learned yet.  
P.S.  Is that James Stewart's book?  I seem to remember this problem from when I took Calculus I.
A: The question is asking for $\frac{d\theta}{dt}$ when $\theta=0$, i.e. when the passing car is right in front of the police car. This works out as
$$\frac{d\theta}{dt}=\frac{110}{20\sec^20}=\frac{110}{20\cdot1^2}=\frac{11}2$$
Note that this is in radians; converting to degrees we get a value of $\frac{11}2\cdot\frac{180}\pi=315.1$ degrees per second.
A: The answer depends on what the words "at that time" mean.
As you found, the rate at which the angle $\theta$ is changing
is a function of the value of the angle $\theta$ at the instant
when you measure the rate of change.
"A car passes $20$ ft from her ... How fast is the angle $\theta$
changing at that time?"
I interpret this to mean that "that time" is the instant when the car
crosses the dotted line in the figure, since that's when the
car "passes" the officer and also when it is $20$ feet away.
Since the car at that time is in exactly the same direction
as the dotted line, the angle $\theta$ is zero.
