Related rates of change involving trig functions 
Question: A train is moving along a straight track at
$75 \text{km h}^{-1}$ heading east. A camera is positioned $2$km north of the train track and is focused on the train.
a) Find the rate of change of the distance between the camera and the train when the train is $4$km from the camera.
b) Find the rate at which the camera is rotating when the train is 4km from the camera. Give your answer in degrees per second correct to the nearest tenth of a degree.

This is a question that has caused my extension maths group a bit of confusion! We have solved part a of the question and we believe the answer is $65$km/h (2sf). This is correct.
Part b is the one we are unsure about. We have set up the problem using a couple of variables and applied various differentiation techniques. The answers we are getting, however, are varied and even our teacher has said that he needs a bit more time to work out which one is correct! I am quite eager to find out whether my technique is correct. The different answers we have got are:

*

*$2.6 \times 10^{-3}$ radians/second (then we can convert this to degrees per second as the question asks)


*$2.6$ degrees/second
I appreciate that by putting this question out there, I may generate different responses, so I will wait to see if a several people start getting the same answer before accepting if it is correct.
Best wishes!
 A: I'll illustrate a general solution to the problem since you seem to be enthusiastic about this topic. Below is a diagram of the situation:

$x$ is the distance of the train to the perpendicular, $d$ is the distance to the observer, $v$ is the train's speed (constant), $s$ is the distance of the train to the observer, and $\theta$ is the angle the train makes with the perpendicular, some number in the open interval $(-\pi/2,\pi/2)$. I'll use dots to denote time derivatives. We know that the train moves at speed $v$, in other words
$$\dot{x}=v$$
We also know, from Pythagoras, that
$$x^2+d^2=s^2$$
So,
$$2s\dot{s}=2x\dot{x}\implies \dot{s}=\frac{x\dot{x}}{s}=\frac{x\dot{x}}{\sqrt{x^2+d^2}}=\frac{xv}{\sqrt{x^2+d^2}}$$
So, if $d=2\text{ km}$, $v=75\text{ km h}^{-1}$ and $x=\pm 4\text{ km}$ we get
$$\dot{s}=\frac{\pm 4\cdot 75}{\sqrt{4^2+2^2}}\text{ km h}^{-1}=\frac{\pm 4}{\sqrt{20}}\cdot 75\text{ km h}^{-1}\approx 67\text{ km h}^{-1}$$
Onto the second part. For this part I'll take positive $x$ on the right. With trigonometry we see that
$$\tan(\theta)=\frac{x}{d}\implies \theta=\arctan\left(\frac{x}{d}\right)$$
Taking time derivatives of both sides we get
$$\sec^2(\theta)\dot{\theta}=\frac{\dot{x}}{d}$$
Now using our expression for $\theta$ and the fact that $\dot{x}=v$,
$$\dot{\theta}=\frac{v}{d~\sec^2\left(\arctan\left(\frac{x}{d}\right)\right)}=\frac{v}{d\left(\frac{x^2}{d^2}+1\right)}=\frac{vd}{x^2+d^2}=\frac{vd}{s^2}$$
I'll leave it as an exercise to prove that
$$\sec(\arctan(t))=\sqrt{t^2+1}$$
Using $v=75\text{ km h}^{-1}$, $s=4\text{ km}$, and $d=2\text{ km}$ we get
$$\dot{\theta}=\frac{75\cdot 2}{4^2}\text{ radians h}^{-1}\approx 0.149\text{ degrees s}^{-1}$$
I don't think either of the provided answers are correct.
A: Firstly, $75 \text{km h}^{-1} = \frac1{48} \text{km s}^{-1}$
Draw a right angled triangle with the triangle 'pointing' right.
Label the top vertex $A$, and then $B$ and $C$ will follow in clockwise fashion.
$AC = 2$km. The rate of change of $BC$ is $\frac1{48}  \text{km s}^{-1}$ and $BC$ is $\frac1{48} t $ km (I assume this comes from speed = distance*time?).
Let $\angle BAC = \theta$.
Then $\tan(\theta) = \frac1{96}t$ and, differentiating both sides with respect to $t$, we get $\frac1{cos^2\theta}\frac {d\theta}{dx} = 1/96$
We can know divide both sides and work out an expression.
We can also work out the value of $\cos \theta$ by plugging in that the hypotenuse is $4$ and so work out $\cos^2 \theta$. Thus, we can plug this value into our expression for $\frac {d\theta}{dt}$, which gives them $1/4 \times 1/96 = 1/384 = 2.6$ degrees per second (?)
Sorry I don't have the complete working.
