Trying to prove the earth is flat

One simple way to get an estimation is to calculate the distance between their heads. $CA' = r + 1.8$ and $CD = r$ , thus $A'D = \sqrt{2*r*1.8 + 1.8^2}$ , which is about 4789. Thus, $A'B'$ is about 9578.

To find precisely the distance, a possibility is to calculate the angle $\theta = ACB$ , which verifies :

$\cos(\frac{\theta}{2}) = \frac{r}{r+1.8} = 0.999999717$ , thus $\theta = 0.000752329733 rad$ . The perimeter $P$ of the Earth is $2\pi r$ , thus $AB = \frac{P.\theta}{2\pi} = 9586$ .

$cos^{-1} (\frac{6371009}{6371010.8}) * 6371009 * 2 = 9578.23$

Let the angle of center of the circle opened to the two people is $2\theta$ , we can obtained

$\cos \theta = \frac{R}{R+1.8}$

Because the angle is almost zero, we can use approximate method, namely Taylor Series

$\frac{R}{R+1.8} \approx 1-\frac{\theta ^2}{2}+ smallterms$

$\theta \approx \sqrt{\frac{3.6}{R+1.8}}$

So the arc $AB$ can be calculated by formula $L = 2R\theta \approx 2R\sqrt{\frac{3.6}{R+1.8}}$

Put in the value of $R=6371009$ to obtain the answer $L \approx 9578.231 \approx \boxed{9500}$

Simply apply Pythagoras' Theorem, $AB=2 \sqrt{(6375009+1.8)^2-6375009^2} \approx 9578.23202239 m$

For those of you approximating cos, I would suggest using sin instead. Since sinx/x converges to 1 very rapidly, you can use x in place of sinx for small x and retain a great deal of accuracy.

The maximum distance that would prevent them from seeing themselves is tangential to the earth surface. So, the angle subtended by arc AB at the center of the earth is 2 times of arccos 6371009/(6371009+1.8) which equals to 0.0861 degrees. So, the length of arc AB is 2 3.1416 6371009*0.0861/360 which equals to 9573.89 metres. So, the correct answer is about 9500 metres.

EZ pathagoras problem don't know why this is on intermediate

Since the earth radius is so large, arc length $\approx2\times\sqrt{(R+1.8)^2 - R^2}=9578.225 \text{ m}.$