A little problem with curvature vs trigonometry

[Kriss3d]

Im working on a little problem for an idea im having.

If you stand on earth and look out over the ocean. Your field of vision would be almost 180 degrees. But lets say 150 degrees for this example.

Being 6 feet above sea level puts your distance to the horizon 3 miles out.

 

This means that by trigonometry you can draw a triangle with 3 miles on two sides and the angle between them being 150 degrees.

 

This would make the line between them be 5.8 miles.

Ive been wondering. By math, how much of earth curvature would be seen on each side of the middle where the curvature from your point of view would be zero ?

Earth will curve down to the sides in respect to the direct front of you. The wider this tangent line the more the curvature will be aparent. Obviously its not possible to register any curvature by looking at the ocean from this close. But math should be able to tell exactly how much it actually is.

 

By my crude paint skills ive made an example. What id like to know is how to get the hight of the tangent to the surface of the curvature when i know that the tangent line is 5,8 miles long. I Just cant quite seem to find the right way to formulate this into the right formula.

[AbstractDreamer]

https://en.wikipedia.org/wiki/Horizon

 

This looks like its got trigonmetry stuff you need