Trigonometry - Horizon Related Word Problem

AI Thread Summary
The discussion focuses on solving a trigonometry problem related to the horizon and the curvature of the Earth. The original poster struggles with understanding the concept of the horizon and the geometry involved, particularly in relation to the Earth's radius and the height of an observer. Key points include the need to visualize the problem correctly, using the radius of the Earth and the height of the observer to determine the distance to the horizon. Participants suggest using the Pythagorean theorem and the tangent-secant theorem for calculations. Understanding the geometry of the situation is crucial for finding the correct solution.
nmnna
Messages
22
Reaction score
3
Homework Statement
Persons A and B are at the beach, their eyes are 5 ft and 6 ft, respectively, above sea level. How
many miles farther out is Person B’s horizon than Person A’s?
Relevant Equations
;;
Hello!
I'm trying to solve this problem.
Here's the diagram I tried to make.
1614415889388.png

I have difficulty understanding this math problem.. I've tried to solve the problem using the symmetry of the triangles but I didn't get the right answer, and I can't seem to understand the "concept" of the horizon here.
So I'll be grateful if you give me some hints.
 
Physics news on Phys.org
If I'm not mistaken, it is necessary to consider the curvature of the earth, which can be considered to be a sphere with a radius of 4000 miles.
 
Yeah, I think your diagram is at least very misleading. You have them looking at like, the side of a mountain or something, which isn't really how the horizon works.
 
You should view it like this. Line BC is of length R, where R is the radius of the Earth. Line AC is of length R+H, where H is the height of the person. Line AB is tangent to the Earth, so line AB is perpendicular to line BC. You want to calculate the distance AB, which is how far the person is seeing. Remember that H<<R, and use approximations.
Horizon.png
 
  • Like
Likes nmnna and Lnewqban
nmnna said:
... but I didn't get the right answer, and I can't seem to understand the "concept" of the horizon here.
So I'll be grateful if you give me some hints.
Welcome, nmnna! :smile:
Did they give you a value of radius to use for this problem?

Please, see:
https://en.m.wikipedia.org/wiki/Earth_radius

As explained above, horizon is where the line of sight of a person hits the interface sky-ocean.
That makes that line of sight a line that is simultaneously tangent to the surface of the ocean (at the horizon) and perpendicular to a line from the horizon to the center of the Earth.

Besides the Pythagorean theorem, you could use the tangent-secant theorem, making the secant line go through the center of the Earth:
https://en.m.wikipedia.org/wiki/Tangent-secant_theorem
z4_6QpKu_EzL8HuhRVT-tVSv65BMHIZ7jR5s4p6-evqZczI1Bg.gif
 
Last edited:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top