Answer
The $\theta $ is decreasing at the rate of $0.1872$ radians/hour.
Work Step by Step
Let C represents the radar station, A represents the plane and $\theta$ represents the angle that the line through the radar station and the plane makes with the horizontal.
We get the figure as shown below.
Since it forms a right triangle, we can use the Pythagoras theorem.
That is, $AC^2=BC^2+AB^2$
Now substitute $AB=6$km
We get, $AC^2=BC^2+6^2=BC^2+36$
Let $AC=y$ and $BC=x$.
We get, $y^2=x^2+36$
Since time $t=12$ min or $t=\dfrac{12}{60}=\dfrac{1}{5}$ hours.
Use speed formula to calculate horizontal distance covered by plane after $t=\dfrac{1}{5}$ hours.
$speed=\dfrac{distance}{time}$
$\implies distance=speed \times time$
$\implies x=800\times \dfrac{1}{5}=160$km
Now substitute $x=160$ in $y^2=x^2+36$ and solve for $y$.
We get, $y^2=(160)^2+36=25600+36=25636$
Now use $\cot{\theta}=\dfrac{base}{perpendicular}$.
We get, $\cot{\theta}=\dfrac{x}{6}$
Now differentiate with respect to time $t$ using the chain rule.
$-\csc^{2}{\theta}\dfrac{d\theta}{dt}=\dfrac{1}{6}\dfrac{dx}{dt}$
Now use $\csc{\theta}=\dfrac{hypotenuse}{perpendicular}$.
We get, $\csc{\theta}=\dfrac{y}{6}$
Now substitute $\csc{\theta}=\dfrac{y}{6}$ in $-\csc^{2}{\theta}\dfrac{d\theta}{dt}=\dfrac{1}{6}\dfrac{dx}{dt}$.
We get, $-\left(\dfrac{y}{6}\right)^{2}\dfrac{d\theta}{dt}=\dfrac{1}{6}\dfrac{dx}{dt}$
Now solve for $\dfrac{d\theta}{dt}$.
$-\left(\dfrac{y}{6}\right)^{2}\dfrac{d\theta}{dt}=\dfrac{1}{6}\dfrac{dx}{dt}$
$\implies -\dfrac{y^2}{6^2}\dfrac{d\theta}{dt}=\dfrac{1}{6}\dfrac{dx}{dt}$
$\implies -\dfrac{y^2}{6}\dfrac{d\theta}{dt}=\dfrac{dx}{dt}$
$\implies \dfrac{d\theta}{dt}=-\dfrac{6}{y^2}\dfrac{dx}{dt}$
Now substitute $\dfrac{dx}{dt}=800$ and $y^2=25636$.
We get, $\dfrac{d\theta}{dt}=-\dfrac{6}{25636}\times 800=-\dfrac{4800}{25636}=-0.1872$ radians/hour
The negative sign shows that the angle is decreasing.
So, the $\theta $ is decreasing at the rate of $0.1872$ radians/hour.
