Please help me find the mistake in my related rates kite problem I have been having utter fits with this assignment. It was due months ago, but because I submitted it on time, I can resubmit until the end of the semester for full points.
A kite 100 ft above the ground moves horizontally. The kite-holder stands still while letting out string at
a rate of 4 ft/sec. The angle made by the string and the ground gradually decreases. At what rate is
this angle changing when 200 ft of string has been let out?"
I calculated this myself, several times using several different approaches, and I always come back to the same answer: $$\frac {d\theta}{dt} = -\frac 1 {100} \frac {rad} {sec}$$ 
This really seems correct! In fact, in class, we did a problem (seemingly) just like this one, but let the rate = 8 ft/sec. And in class, we determined that the rate was  $$\frac {d\theta}{dt} = -\frac 1 {50} \frac {rad} {sec}$$ 
So, concerning the one at hand, my study buddy got a grade back on it, and the teacher apparently said that he did it incorrectly, BUT got the right answer somehow. The correct answer is, supposedly,
$$\frac {d\theta}{dt} = \frac {\sqrt 3}{50} \frac {rad} {sec}$$
?!?!?!?!?!
I even asked this question on Chegg, for Pete's sake, and I got the exact same answer as my own! Where does this newfangled answer even come from??
Is there something blantantly obvious I'm missing here? I mean... maybe?? But what?
Please confirm or discredit my insanity. I don't know what to believe because I don't know how to believe anymore.
Cheers,
-Jon
 A: Funny, I get yet a third answer:
Let $l$ be the distance from the person to the kite. Initially $l=100$ and increases at $4$ feet per second.
The relationship between the angle $\theta$ and $l$ is
$$\sin \theta = \frac{100}{l}.$$
Therefore
$$\frac{dl}{dt}\sin \theta + l\cos\theta \frac{d\theta}{dt} = 0.$$
Now after $200$ feet of string has been released, $l=300$ and $\sin\theta=\frac{1}{3}$. We can compute $\cos\theta$ by solving the triangle with leg $100$ and hypotenuse $300$: $\cos\theta = \frac{\sqrt{300^2-100^2}}{300} = \frac{2\sqrt{2}}{3}.$ Therefore
$$\frac{d\theta}{dt} = -4\cdot \frac{1}{3} \cdot \frac{1}{300\cdot \frac{2\sqrt{2}}{3}} = -\frac{\sqrt{2}}{300}.$$
A: $$\sin\theta=\frac{100}{l}$$
$$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$
$$\frac{\sqrt{3}\times 100 }{200}.\frac{d\theta}{dt}=-\frac{1}{100}$$
$$\frac{d\theta}{dt}=-\frac{1}{50\sqrt{3}}\text{rad per sec}$$
A: Given The Dead Legend's calculations, 
$$\sin\theta=\frac{100}{l}$$
$$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$
$$\frac{\sqrt{3}\times 100 }{200}.\frac{d\theta}{dt}=-\frac{1}{100}$$
$$\frac{d\theta}{dt}=-\frac{1}{50\sqrt{3}}\text{rad per sec}$$
I finally saw what I did wrong: I was taking
$$cos\theta = \frac {\sqrt 3}{2} $$
and not
$$cos\theta = \frac {100\sqrt 3}{200} $$
Which makes sense now.
However, I still don't think the answer is right.
dl/dt = 4 ft/sec, l = 200. I followed The Dead Legend's calculations and got back
$$\frac{d\theta}{dt} = - \frac{\sqrt 3}{150}$$
I plugged it into a calculator and got back the same results.

So how did my study buddy get the right answer? The only possible answer is that he didn't. My teacher must have thought he did.
When I reviewed my buddy's notes, he followed the same approach as The Dead Legend. 


*

*He derived $$\sin\theta=\frac{100}{l}$$

*He got back 
$$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$


And everything went swimmingly (except for a dropped minus sign), where he got the equation down to 
$$\frac {400}{40000} \cdot \frac {200}{100 \sqrt 3}$$
Which could be simplified to 
$$\frac {1}{100} \cdot \frac {200}{100 \sqrt 3}$$
Then he simplified 200/100√3.
Here is where he went wrong.
$$\frac {200}{100 \sqrt 3}$$
$$= \frac {2}{\sqrt 3}$$
$$= 2 \sqrt 3$$
Oops.
When I rationalized the denominator, I got back
$$= \frac {2 \sqrt 3}{3}$$
And from THERE, I finished his calculation (despite the curiously absent minus sign.)
$$\frac {1}{100} \cdot \frac {2 \sqrt 3}{3}$$
$$= \frac {2 \sqrt 3}{300}$$
$$= \frac {\sqrt 3}{150}$$
Tack on the dropped minus...
$$= - \frac {\sqrt 3}{150}$$
There it is, folks. The elusive answer, if indeed everything was done correctly.
One way or another, my brain hurts. So glad I'm getting a different teacher for Calc II.
