Solving for the radius of the Earth based on distance to horizon problem Lets say you are standing on the top of hill with height $d$ and from the top of this hill you can see the very top of radio tower on the horizon. Your goal is to determine the radius of the Earth, so you drive from the base of the hill, in a straight line (along an arc) to the base of the radio tower. Your odometer measures an arc length, $s$ and you can measure the height of the radio tower, $a$. So, what is $R$?

I may have just forgotten my basic algebra but I am trying to solve this by using three equations and three unknowns:
$$s = R(\phi_1+\phi_2) $$
where
$$ \cos(\phi_1) = \frac{R}{R+a} $$
and 
$$\cos(\phi_2) = \frac{R}{R+d}.$$
Here, $a$, $d$ and $s$ are all known. So we have three equations and three unknowns where $R$ is the radius of the Earth to be determined, $\phi_1$ and $\phi_2$ are the angles of their respective arc lengths. 
How does one solve this?
As a note, this is not a homework problem. I actually want to go perform this experiment to see how accurate my determined radius is.
 A: $$
R = \underbrace{\frac a {\sec\varphi_1 - 1} = \frac d {\sec\varphi_2 - 1} = \frac s {\varphi_2+\varphi_2}} \tag 1
$$
I think this will have to be done numerically rather than in closed form.
Let's try a first approximation using $\sec\theta-1 \approx \dfrac{\theta^2} 2$:
$$
\frac a {\varphi_1^2/2} = \frac d {\varphi_2^2/2} = \frac s {\varphi_2+\varphi_2}
$$
$$
\frac{\varphi_1^2}{2a} = \frac{\varphi_2^2}{2d} = \frac{\varphi_1 + \varphi_2} s \tag 2
$$
Clearly $\varphi_1=\varphi_2=0$ is a solution of both $(1)$ and $(2)$, but we seek non-zero solutions.  Letting $\alpha=\varphi_2/\varphi_1$ is licit because $\varphi_2\ne0$.  Then we have
$$
\alpha^2 = \frac d a \quad \text{and so} \quad \frac {\varphi_1(1 + \sqrt{d/a\,{}})} s = \frac{\varphi_1^2}{2a}.
$$
$\phantom{0}$
\begin{align}
\varphi_1 = \frac{1 + \sqrt{d/a\,{}}} s \cdot 2a & = 2(\sqrt a + \sqrt d) \cdot \frac {\sqrt a} s \\[12pt]
\varphi_2 & = 2(\sqrt a + \sqrt d) \cdot \frac {\sqrt d} s.
\end{align}
If $\varphi_1,\varphi_2$ are small, this should be quite close to the answer.
A: Treating each side of the symmetric situation separately, we have $\cos\phi_1=R/(R+a)$, and thus for $\phi\ll1$ (which holds for realistic hills and radio towers) $1-\frac12\phi_1^2\approx R/(R+a)$. Solving for $\phi_1$ yields $\phi_1\approx\sqrt{2a/(R+a)}\approx\sqrt{2a/R}$. Likewise, $\phi_2\approx\sqrt{2d/R}$. Then $s=R(\phi_1+\phi_2)\approx\sqrt{2R}(\sqrt a+\sqrt d)$, so
$$
R\approx\frac{s^2}{2\left(\sqrt a+\sqrt d\right)^2}\;.
$$
