Using Trigonometry to Find Distance from a Flagpole on a Hill

Click For Summary
A woman on a hill observes a 60 ft tall flagpole, with an angle of depression to the bottom at 14 degrees and an angle of elevation to the top at 18 degrees. By using trigonometric functions, she establishes two right triangles to calculate her distance from the pole, denoted as x. The equations derived from the tangents of the angles lead to the formula x = 60 / (tan 18 + tan 14), resulting in a distance of approximately 104.485 ft. The discussion also touches on the treatment of significant figures, noting that for this problem, the answer can be presented as 104 ft or 104.5 ft. Overall, the problem illustrates the application of trigonometry in real-world scenarios.
LordofDirT
Messages
15
Reaction score
0
1. A woman standing on a hill sees a flagpole that she knows is 60 ft tall. The angle of depression to the bottom of the pole is 14 degrees, and the angle of elevation to the top of the pole is 18 degrees. Find her distance x from the pole.

2. I draw a diagram of the problem with a person on a hill looking straight out to the flagpole. This line I labeled x. I then drew an angle of elevation of 18 degrees to the top of the flagpole. and one of 14 to the bottom of the flagpole.

I made y=opposite leg of 18 degrees, and z=opposite leg of 14 degrees.


I know that y+z=60, y=60-z, and z=60-y


All I need to figure out is how to get either y or z, and then the problems a sinch.
 
Physics news on Phys.org
You almost have it,
Consider it as two separate right triangles.
Now find the height of the unknown segments of the pole ( from x-top and x-bottom ) in terms of the angle and X.
You also know that these two distances sum to be 60 so you should end up with two equations in only X that add to 60ft, then it's simple!

Post back if you need more details
 
think i got it...

I took the tangent of 18 degrees and ended up with the equation y=xtan18 for the leg of the top triangle, and the tangent of 14 degrees to get z=xtan14.

So I figured...

xtan18 + xtan14 = 60 to find x

x(tan18 + tan14) = 60

x=60/(tan18 + tan14)

x=104.485

2 sig figs so the distance should be 104 ft?
 
LordofDirT said:
x=104.485 ft.

That's fine!

2 sig figs so the distance should be 104 ft?

Well, 104 wouldn't be 2 sig-figs... In math problems of this sort, you don't really have to worry too much about significant figures. The angles, such as 18º, are taken to be exact, and so have unlimited significant figures. (The flagpole height of 60 ft. is ambiguous: that could be as few as one sig-fig, but you would be all right in taking it to be exact as well...) So you would be safe in giving the answer as 104 ft. or even 104.5 ft.

If this were a physics problem (or one where strict observation of significance had to be observed), we should state the answer as 1.0·10^2 feet for the two significant figures.
 
Thanks!
 
mgb_phys said:
You almost have it,
Consider it as two separate right triangles.
Now find the height of the unknown segments of the pole ( from x-top and x-bottom ) in terms of the angle and X.
You also know that these two distances sum to be 60 so you should end up with two equations in only X that add to 60ft, then it's simple!

Post back if you need more details

You know I think you were teating it like two right triangles. I think you were doing just fine. The thing that was implied but not said in the problem was that the point on the flagpole that was horizontal was known. What if it was not, but we knew that the angle from top to bottom was 32 deg.? You got me thinking. Very good post. Thanks.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
11
Views
6K
  • · Replies 8 ·
Replies
8
Views
9K
Replies
1
Views
2K
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K