Related rates problem. Determine the speed of the plane given the following information. 
I started by noting that I am looking for $\displaystyle\frac{dx}{dt}$ and I am given $\displaystyle\frac{d\theta}{dt}=1.5^{\circ}/\mbox{s}$.
I then related $\theta$ to $x$ by $\displaystyle\cot(\theta)=\frac{x}{6000\mbox{m}}$, then taking their derivatives with respect to $t$, I get $\displaystyle\frac{dx}{dt}=-6000\mbox{m}\csc^2(\theta)\frac{d\theta}{dt}$, but i'm not sure if this is correct, this number seems too big, and it is also negative. Thoughts?

The side $x$ of the triangle is shrinking, while $X = x_0 - x$ is increasing, with $v = \dot{X} = - \dot{x}$, so the minus sign of $\dot{x}$ is OK.
$$
\tan 60^\circ = \sqrt{3} = \frac{6000\mbox{m}}{x_0}
\iff x_0 = 3464 \mbox{m}
$$
With
$$
\tan \theta = \frac{6000 \mbox{m}}{x} \Rightarrow 
x = 6000 \mbox{m}\frac{\cos \theta}{\sin \theta}
$$
$$
v = -\dot{x} = 6000 \mbox{m}\frac{\dot{\theta}}{\sin^2 \theta}
$$
thus when $\theta = 60^\circ$ we have $\sin 60^\circ = \sqrt{3}/2$ and
$$
v = 6000 \mbox{m}\frac{(3/2^\circ /\mbox{s}) (\pi/180^\circ)}{3/4}
= 6000 \mbox{m}\frac{((\pi/120) 1/\mbox{s})}{3/4}
= \frac{200}{3} \pi \frac{\mbox{m}}{\mbox{s}}
\approx 209 \frac{\mbox{m}}{\mbox{s}}
$$
