Answer
(a) $\theta=cos^{-1}(\frac{3960}{h+3960})$
(b) $s=7920\cdot \theta$
(c) $s=7920\cdot cos^{-1}(\frac{3960}{h+3960})$
(d) $1761$ mi.
(e) $197$ mi.
Work Step by Step
(a) Use the figure given with the problem and use $R$ for the radius of the earth,
we have $cos\theta=\frac{R}{R+h}=\frac{3960}{h+3960}$ and $\theta=cos^{-1}(\frac{3960}{h+3960})$
(b) Clearly $\frac{s}{2}=R\theta$ ($\theta$ in radian) and we have $s=2\times3960\times\theta=7920\cdot \theta$
(c) Combine the results from (a) and (b), we have $s=7920\cdot cos^{-1}(\frac{3960}{h+3960})$
(d) Given $h=100$, use the relationship in (c) to get $s=7920\cdot cos^{-1}(\frac{3960}{100+3960})\approx1761$ mi
(e) Given $s=2450$, we have $2450=7920\cdot cos^{-1}(\frac{3960}{h+3960})$ which leads to
$\frac{3960}{h+3960}=cos(\frac{2450}{7920})=0.9525$ and $h=\frac{3960}{0.9525}-3960\approx197$ miles
