Rate at which $x$ changes with respect to $\theta$ 
A ladder $10$ft long rests against a vertical wall. Let $\theta$ be the angle with the horizontal. The top of the ladder is $x$ feet above the ground. If the bottom of the ladder is pushed toward the wall, find the rate at which $x$ changes with respect to $\theta$ when $\theta=60$ degrees. Express the answer in units of feet per degree.

So I start by illustrating an imagine of a ladder against a building to make a triangle where hypotenuse $=10$, $x$ is the vertical, and $\theta$ is the angle opposite the vertical. So,
$$\sin \theta=\frac{x}{10} \implies x=10\sin \theta$$
And we need to find the rate at which $x$ changes with respect to $\theta=60$ degrees,
$$x'=10\cos \theta \implies x'(\theta)=5$$
Question:
Does this show that the rate at which $x$ changes with respect to $\theta=60$ degrees is equal to $5$? i.e For $\theta=60$ degrees, $\frac{dx}{d\theta}10\sin \theta=10\cos \theta=5$. If so, how can I be sure that my units are in $\frac{feet}{deg}$?
 A: Well, $\theta$ is in degrees, so the derivative of $\sin(\theta)$ is not $\cos(\theta)$ but $\frac{\pi}{180}\cos(\theta)$. This is because: 
$$\frac{d}{dx}[\sin(\theta)] = \cos(\theta)$$ for radian values of $\theta$ only.  
To get degree values: 
$\frac{d}{dx}[\sin(\theta)]$ with $\theta$ in radians is equal to $\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)]$ with theta in degree values but the sine function  still taking radian values. Therefore, the new derivative is: 
$$\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)] = \frac{\pi}{180}\cdot \cos(\frac{\pi}{180}\cdot \theta)$$ 
where theta is in degrees but cosine is still taking radian values. Therefore, if we change cosine and sine to take degree values we can simplify to: 
$$\frac{d}{dx}[\sin(\theta)] = \frac{\pi}{180}\cdot \cos(\theta)$$  for degree values of $\theta$
Therefore, with your previous evaluation, I believe the correct answer is actually $\frac{\pi}{36}$. If I assume that as your answer, I can answer your other questions: 
For your first question, we can be sure of that because $x'$ is equal to: $\frac{dx}{d\theta}$, which is read as the rate in change of $x$ with respect to $\theta$. 
For your second question, $x$ is already in feet and $\theta = 60$ degrees (as stated in the question), so all you have to do is state it like this: 
The rate of change of $x$ with respect to $\theta$ at $\theta$ = 60 degrees is $\frac{\pi}{36}$ feet per degree. 
