Answer
a.) See work to get to $h = r(\csc(\theta) - 1)$
b.) $-13,717.8424$ mph
Work Step by Step
a.) Notice how a right triangle is formed in the picture so that $sin(\theta) = \dfrac{opposite}{hypotenuse}= \dfrac{r}{r + h}$
By isolating h in this equation, we can find $h = r(csc(\theta) - 1)$
$(sin(\theta))^{-1} = (\dfrac{r}{r + h})^{-1}$
$csc(\theta) = \dfrac{r + h}{r}$
$r \cdot csc(\theta) = r + h$
$rcsc(\theta) - r = h = r(csc(\theta) - 1)$
b.) Take the derivative of both sides of $h = r(csc(\theta) - 1)$ and then plug in the given values.
$\dfrac{d}{d\theta}(h) = \dfrac{d}{d\theta}(r(csc(\theta) - 1))$
$\dfrac{dh}{d\theta} = r(-csc(\theta)cot(\theta))(\dfrac{d\theta}{d\theta}) + (csc(\theta) + 1)(\dfrac{dr}{d\theta})$
GIVEN:
$\theta = 30º $
r = 3960 miles
Since the radius of the earth (r) is not changing, $\dfrac{dr}{d\theta}$ is zero.
$\dfrac{dh}{d\theta} = -3960(csc(30)cot(30))(1) + (csc(30) + 1)(0) = -3960(csc(30)cot(30))$
$\dfrac{dh}{d\theta} = -13,717.8424$ miles per hour
NOTE: Make sure that your calculator is in degree mode since we are using 30º
