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### The Top of the Tower

```
Date: 07/20/2001 at 15:02:38
From: Ashley Sherman
Subject: Word problem

Given a 300-mile-tall tower on the earth, how far away from it can you
go before the top disappears below the horizon? (Assume unlimited
visibility.)

We've drawn diagrams and triangles to try to figure it out using
trigonometry, but we're looking for an equation(s) that will help us
```

```
Date: 07/23/2001 at 11:41:00
From: Doctor Ian
Subject: Re: Word problem

Hi Ashley,

It's hard to draw a circle with a keyboard, but let's pretend this is
the earth:

.
.       .
.           .

.             .

.           .
.       .
.

Also, let's pretend the earth is a sphere, and that it's perfectly
smooth, even though neither of those things is true.

If your eye is at ground level at some point, then you can look out
tangent to the earth:

<----------------
.       .
.           .

.             .

.           .
.       .
.

If the top of a tower of height h just intersects your line of sight,

<---------.------
.       .   .
.           .   h

.             .

.           .
.       .
.

then you can form a right triangle with you, the center of the earth,
and the tip of the tower at the corners:

A          C
<---------.------
.   .   .   .
.     .     .   h
.  .
.      .      .
B
.           .
.       .
.

If the radius of the earth (in miles) is r, then

AB = r

BC = r + h

and

r
cos(ABC) = -----
r + h

On a sphere, the distance along the surface (s) corresponding to a
central angle theta is

s = r * theta

So the distance you'd have to walk from point A to the base of the
tower would be

s = r * theta

r
= r * arccos( ----- )
r + h

For r = 3950 miles and h = 300 miles,

3950
s = 3950 * arccos( ---- )
4250

= 1493 miles

That seems like a pretty long distance! Is it reasonable? It's tough
to visualize something 300 miles tall. Let's think about something
more realistic, like the Sears Tower.

The Sears tower is 1450 feet tall, which works out to be about 0.275
miles. So our equation says that we should be able to see it up to

3950
s = 3950 * arccos( -------- )
3950.275

= 46.6 miles

which means that you would just be able to see the Sears tower from
the Indiana Dunes State park - if you were lying on the ground. I know
for a fact that you can see it from there standing up, because I grew
up near there.

Okay, so what if you're _not_ lying on the ground?

ground, what's the farthest point you can see?

A'
| h'
|  .
.   .   .
.     .     . C'
.  .
.      .      .
B
.           .
.       .
.

Now we have a different right triangle (A'C'B is the right angle), and
we know that

r
cos(A'BC') = ------
r + h'

This looks familiar, doesn't it?

Now extend the line A'C', and you have the same problem we just
solved. The central angle is now (A'BC' + ABC), so

s = r * (A'BC' + ABC)

r                  r
= r *[arcos( ------ ) + arccos( ----- )]
r + h'             r + h

Let's check that again for the Sears Tower. If you're 6 feet tall,
that's about 0.0011 miles tall, so

3950                  3950
s = 3950 * [arccos( --------- ) + arccos( -------- )]
3950.0011             3950.275

= 49.6 miles

which isn't a lot of difference.

Now, here's the interesting thing. From Indiana, it seems as though
you can see, not just the top of the building, but the top half. The
greatest distance from which that should be possible is

3950                  3950
s = 3950 * [arccos( --------- ) + arccos( --------- )]
3950.0011             3950.1375

= 35.9 miles

I just checked with maps.yahoo.com, and the driving distance from the
house where I grew up in Indiana (just outside the Dunes State Park)
to the Sears Tower is 49.7 miles, and the distance from the Dunes
State Park to the Sears Tower is a little less than 40 miles (as
measured on the map that accompanies the driving instructions). So
this appears to work out quite well.

more, or if you have other questions.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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