Height of Flagpole Problem

geekily

Junior Member
Joined
Jan 24, 2007
Messages
93
A boy and his friend wish to calculate the height of a flagpole. One boy holds a yardstick vertically at a point 40 feet from the base of the flagpole. The other boy backs away from the pole to a point where he sights to top of the pole over the top of the yardstick. If his position is 1 foot 9 inches from the yardstick and his eye level is 2 feet above the ground, find the height of the flagpole.

I'm really not sure how to start with this one. I remember doing "height of ____" problems in algebra, but never like this. I drew a diagram, and marked the angle at the flagpole and the ground 90º. Now, I can take a protractor and measure the angle of the top of the flagpole and the line from his line of sight, but I can't make it perfect because part of the problem is near the binding part of the page, so the diagram is curved and I can't know if my picture is exactly perfect. Even if I did know, I'm not quite sure what good it would do me. Any help would be greatly appreciated. Thank you!
 
Assuming that the yardstick is resting on the ground and that the boy doing the eyeballing is squating down, you have two nested right trianges.

The base of the overall triangle is the horizontal line-of-sight, from the boy to the point on the flagpole which is two feet off the ground. The taller triangle's height, "h", is the flagpole (less those initial two feet); the shorter height, "1", is the yardstick (being the three feet in one yard, less the two feet above ground that don't count). Between the flagpole and the yardstick you have a length of "40"; between the yardstick and the boy you have a length of "1.75".

Use similar triangles to find the value of "h". Then add the two feet back in to get the height. :wink:

Eliz.
 
geekily said:
A boy and his friend wish to calculate the height of a flagpole. One boy holds a yardstick vertically at a point 40 feet from the base of the flagpole. The other boy backs away from the pole to a point where he sights to top of the pole over the top of the yardstick. If his position is 1 foot 9 inches from the yardstick and his eye level is 2 feet above the ground, find the height of the flagpole.

I'm really not sure how to start with this one. I remember doing "height of ____" problems in algebra, but never like this. I drew a diagram, and marked the angle at the flagpole and the ground 90º. Now, I can take a protractor and measure the angle of the top of the flagpole and the line from his line of sight, but I can't make it perfect because part of the problem is near the binding part of the page, so the diagram is curved and I can't know if my picture is exactly perfect. Even if I did know, I'm not quite sure what good it would do me. Any help would be greatly appreciated. Thank you!
Code:
     F
      |\
      |  \
      |    \
      |      \
      |        \Y
      |        | \
      |        |   \ 
      |________|_____\
      O       O'     B

OF = flagpole

O'Y = Yardstick = 3'-2' = 1'

OO' = 40'

O'B = 1'9" = 1 3/4' = 7/4'

Now use stapel's suggestion to solve the problem.
 
Oh, thank you both so much! Using similar triangles definitely makes sense since we've been doing a lot of those. I'm still having a bit of trouble, though.

I made O'B = 7/4, and O'Y = 1'. (but I changed O to A so I wouldn't get confused with all the Os.) Then I did Pythagorean Theorem and came up with square root of 4.0625 for BY. I proved triangle ABY similar to OBF, then set them equal to each other: AB/OB = BY/BF = AY/OF. This turned into 1.75/41.75 = sq. root(4.0625) / BF = 1/OF. Now I figured I could use that first ratio to find the others. However, setting 1.75/41.75 = sqrt(4.0625)/BF didn't pan out because cross-multiplying gave me 1.75x/89.56810345, which is not the same ratio. I tried 1.75/41.75 = 1/OF, but of course, that just gave me the same numbers when I cross-multiplied. What am I doing wrong?

Thank you so much for all your help! I don't know what I'd do without it.
 
You need to find OF - correct?

Use

OF/O'Y = OB/O'B

OF = OB/O'B * O'Y ........................now finish it
 
Um, okay...

So OF/O'Y = OB/O'B is OF/1 = 41.75/1.75

and OF = OB/O'B * O'Y is OF = 23.85714286

Is that right?
 
geekily said:
Um, okay...

So OF/O'Y = OB/O'B is OF/1 = 41.75/1.75

and OF = OB/O'B * O'Y is OF = 23.85714286

Is that right?

You tell me....

What did the problem ask?

and

What did you find?
 
Well, the problem asked the height of the flagpole, which is OF, so I believe that's what I found. I just don't know if I found the right answer. (That is to say, I don't know if I was correct with the numbers I plugged in and such.)
 
Read the question carefully - and study the diagram and the assumptions.

You have not found the height of the flagpole - not yet. You are very close though....
 
My guess would be that my error has something to do with "If his eye level is 2 feet above the ground", but I'm not sure why since I subtracted 2 from the yardstick to begin with. Should I wait until the end to do that or something?
 
geekily said:
I'm not sure why since I subtracted 2 from the yardstick to begin with.
According to the exercise, the boy's line-of-sight is two feet above the ground. If you don't start from his height, how would you get an horizontal base to the triangles?

Eliz.
 
Okay, I guess that makes sense, but I thought I was supposed to use "O'Y = Yardstick = 3'-2' = 1'" like Subhotosh Khan said. Gah, I'm so confused. Maybe it'll make more sense to me in the morning.

Thank you guys so much for your help. :)
 
geekily said:
I thought I was supposed to use "O'Y = Yardstick = 3'-2' = 1'" like Subhotosh Khan said.
If the yardstick (three feet long) is resting on the ground, and you're measuring things two feed above ground, what else would the height of the stick be? :shock:

Eliz.
 
Is it that it would either add or subtract one foot to the height of the flagpole? (Assuming that we already did 3-2?) I *think* it would subtract one foot from the height, meaning whatever I came up with for the height of the flagpole, I would have to add 1.

Before, then, I came up with an answer of 23.85714286. If I add one to that, then, the flagpole = 24.85714286. Is that right?

Thank you again for your help. You have no idea how much I appreciate it.
 
geekily said:
Is it that it would either add or subtract one foot to the height of the flagpole? (Assuming that we already did 3-2?) I *think* it would subtract one foot from the height, meaning whatever I came up with for the height of the flagpole, I would have to add 1.

Before, then, I came up with an answer of 23.85714286. If I add one to that, then, the flagpole = 24.85714286. Is that right?....NO - think again carefully

Also have you ever thought - how would you measure 24.85714286 ft - even the LASER measurement devices are not that accurate.


Thank you again for your help. You have no idea how much I appreciate it.
 
I have thought carefully, I swear, I really have. I turned it in this morning, so I guess it doesn't matter, anyway. As for the 24.85714286 ft, I thought about how it would be impossible to measure that as well, but my teacher always takes off points if we round, so I figured it was safest to include it all.

Thanks for your help, anyway.
 
I would have written it as:

height = 24.85714286 ft = ~ 24' 10"
 
Well, it would definitely make more sense to write it your way. :)

I guess it's just good to know that I was able to come close, then. Thanks!
 
Subhotosh Khan said:
Revised picture
Code:
     F
      |\
      |  \
      |    \
      |      \
      |        \Y
      |        | \
      |        |   \ 
      |________|_____\
     O|      O'|     | B
      |________|_____|
     G        G'       G"

OG = flagpole

O'Y = Yardstick = 3'-2' = 1'

OO' = 40'

O'B = 1'9" = 1 3/4' = 7/4'

OG = O'G' = BG" = 2'

GG" is the ground level

Now use stapel's suggestion to solve the problem.
 
Oooh, so I need to add back *two* feet, then? So, 23.85714286 + 2 = 25.85714286 (or about ~ 25' 10")?
 
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