Rate of change in length of hypotenuse I am absolutely stumped here. I got this problem on my Calc 1 exam and have no clue how to solve it. 
We are given a triangle like this: 
https://i.imgur.com/Rn9SIhv.jpg (Sorry I can't post images)
Then we are told that when θ = π/6, x = 20
So, I have sin θ = 10/x = 1/2
We are asked to find the change in x when θ changes by -0.05 radians. We are not allowed to use a calculator.
Any help with this would be hugely appreciated.
 A: Check your question. Are you sure it's a rate of change you're asked to find for $x$ rather than simply a change for $x$?
Assuming what they're asking for is the estimate of a small change of $x$ for a small change of $\theta$ of $-0.05$ radian, you would work it out like so:
The initial conditions help you verify that the triangle is right (you can't assume from a drawing).
$x = \frac{10}{\sin\theta}$
$\frac{dx}{d\theta} = - \frac{10\cos\theta}{\sin^2\theta}$
For the approximation of small changes, you can rearrange to:
$\delta x \approx - \frac{10\cos\theta}{\sin^2\theta} \delta \theta$
and if you're given $\delta\theta = -0.05$ radian, you get:
$\delta x \approx - \frac{10\frac{\sqrt 3}{2}}{(\frac 12)^2} \delta \theta = - \frac{10\frac{\sqrt 3}{2}}{(\frac 12)^2} (-0.05) = \sqrt 3$
The answer would be the same numerically if you had been given the rate of change of $\theta$ in terms of radian per time, e.g. $\frac{d\theta}{dt} = -0.05 rad/sec$ but there, the chain rule based equation you'd be using would be $\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$ which would give the same numerical value (except the units would be in terms of length over time and this would be exact, not an approximation).
