Converting the Great Circle distance to direct distance between two points on earth? Apologies if this question has been asked before.
Across the surface of the Earth, the distance between London and New York is 5567 km. Given that the earth has a radius of 6371 km, what is the distance between London and New York, supposing a path between them was dug through the earth's crust?
I am interested because I would like to know the minimum theoretical latency for data transmission between the two cities (given the speed of light), but I don't know how to calculate it myself. (I made an attempt and got 9770km, which is obviously wrong...)
 A: This means $\theta$, your angle between lines extending from Earth's centre to London and New York, is equal to $\frac{5567}{6371}=0.8738$ radians. 
Now let's construct a triangle with your proposed tunnel, the radius from Earth's centre to NY, and from Earth's centre to London. The angles in a triangle add to $\pi$ radians, and since two of the legs of the triangle (the two radii) are the same, we know the two remaining angles are the same. Thus, $\frac{\pi-0.8738}{2}=1.1339$ is the measure of the other two angles.
Using law of sines and letting $x$ be the length between NY and London:
$$\frac{\sin1.1339}{6371}=\frac{\sin0.8738}{x}$$
$$x=\frac{6371\sin0.8738}{\sin1.1339}=5391.5475$$
So the distance between New York and London is $5391.5475$ km.

A: If you want a simple function that will provide the answer given the great circle distance as input, the following is adequate:
def through_crust_distance(sphere_dist):
    return ( 6378*sin(sphere_dist/6378) ) / sin( (math.pi-(sphere_dist/6378)) / 2 ) 

A: We assume that Earth is a sphere of radius $r$ and consider an intersection of it to help our intuition. 
Let $A,B$ be two points on the surface of the Earth. Then we can consider them as points that lie on a circle with radius $r$. The distance between them is the arc length $\theta r$, where $\theta$ is the angle $AKB$, where $K$ denotes the center of the Earth, and thus the center of the circle. 
We are interested in evaluating the direct distance $AB$. This is equal to the length of the chord $AB$. Let it be equal to $c$.
Since $AKB$ is an isosceles triangle, by the law of cosines we have that 
$$\cos \theta=1-\frac{c^2}{2r^2}\Rightarrow\\c^2=2r^2(1-\cos\theta)$$
Since we know $\theta,r$ we evaluate $c$.
A: You don’t need the Law of Sines or the Law of Cosines.
If the vertex angle is $\theta$, just draw the perpendicular bisector, and see that the base of your isosceles triangle is $2r\sin(\theta/2)$.
