Derivatives problem - what am I doing wrong? The problem:

A shuttle lifts off vertically from a point $4$ miles from the command post.  For the first $20$ seconds of flight, its angle of elevation changes at a constant rate of $3$ degrees per second.  What is the velocity of the shuttle when the angle of elevation is $30$ degrees?

I made a right triangle picture and realized that $\tan \theta = \frac x4$.  This means that $ x=4\tan \theta$ which is the same as $ x=4\tan(3t)$ because $\theta=3t$ if the angle is changing at $3$ degrees per second.

The velocity will be $\frac {dx}{dt}$ so differentiating the equation gives $\frac {dx}{dt}=12\sec^2 (3t)$. Substituting back gives me $\frac {dx}{dt}=12\sec^2\theta$.  

Finally, the problem asks for the velocity when the angle of elevation is $30$ degrees, which is $\frac \pi 6$, so evaluating my derivative gives $12\sec^2 \frac \pi 6$ which is $16$.

The text says the answer is supposed to be $\frac {4\pi} {45}$, and I can't see my error.  Please help.
community wiki answer so that the post can be closed.

To get the correct answer, you have to express the rate at which the angle is changing in radians per second.  Since $3^\circ$ is equivalent to $\pi/60$ radians, 
$$x = 4\tan\left(\frac{\pi}{60}t\right)$$
{d(θ)/dt} = 3 degrees/second

θ = 30 degrees

tan θ = (x/4)

(dx/dt) = 4 sec^2 (θ) (dθ/dt)

(dx/dt) = 4 sec^2 (30 degrees) (3 degrees/sec)

(dx/dt) = 12 degrees/{cos^2(π/6) second --- change 30 degrees to radian = (π/6)
                                        --- change sec θ to (1/cos θ)

(dx/dt) = 12 degrees/(3/4)second

(dx/dt) = 16 degrees second

Then change 16 degrees to radian which is 16 degrees times (π/180 degrees)

Now for your final answer,

(dx/dt) = (4π/45) second