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Hard question on Similar Triangle and Trigonometry

Masaru

New member
I have a problem with the following question:

Two surveyors estimate the height of a nearby hill. One stands 5 m away from the other on horizontal ground holding a 3 m stick vertically. The other surveryor finds a "line of sight" to the top of the hill, and observes this line passes the vertical stick at 2.4 m. The distance from the top of the stick to the top of the hill is 1500 m (as measured by laser equipment). How hight, correct to the nearest metre, is their estimate of the height of the hill?

The text says that the answer is 720 m, but I have no idea how to get to this answer.

This question is for Year 10 (junior high school level), so there should be an easy way to work it out using Similarity and/or Pythagoras.

Please find my diagram attached here so you can follow my working below:

The triangles CEF and CBA are similar.
So CE/CB = EF/BA
So (x-2.4)/2.4 = [squre root of (1500^2 - (x - 3)^2)]/5

However, when I used an online graphic calculator, I got the answer 652 m, not 720 m.

Also Year 10 students are not expected to work out this kind of complex quadratic equation let alone using a graphic calculator.

I would much appreciate it if someone can tell me what is wrong with my working (or understanding).

Thank you.
 

Attachments

I have a problem with the following question:

Two surveyors estimate the height of a nearby hill. One stands 5 m away from the other on horizontal ground holding a 3 m stick vertically. The other surveryor finds a "line of sight" to the top of the hill, and observes this line passes the vertical stick at 2.4 m. The distance from the top of the stick to the top of the hill is 1500 m (as measured by laser equipment). How hight, correct to the nearest metre, is their estimate of the height of the hill?

The text says that the answer is 720 m, but I have no idea how to get to this answer.

This question is for Year 10 (junior high school level), so there should be an easy way to work it out using Similarity and/or Pythagoras.

Please find my diagram attached here so you can follow my working below:

The triangles CEF and CBA are similar.
So CE/CB = EF/BA
So (x-2.4)/2.4 = [squre root of (1500^2 - (x - 3)^2)]/5

However, when I used an online graphic calculator, I got the answer 652 m, not 720 m.

Also Year 10 students are not expected to work out this kind of complex quadratic equation let alone using a graphic calculator.

I would much appreciate it if someone can tell me what is wrong with my working (or understanding).

Thank you.
Unless I am making the same mistake as you, I agree with your answer of 652m. (Textbooks have been known to be wrong!)

The equation simplifies to a quadratic equation when you square both sides so it could be solved by a Year 10 student (using the quadratic formula).
 

Otis

Member
I agree with the cat. Probably a typo, somewhere.

I worked it bit differently (\(\displaystyle x = \overline{DE}\) ) but also got 652m, rounded.

Did you notice the surveyor taking "line of sight" has their eyeballs on the ground? Were they to move closer to the vertical stick (4.41m away, instead of 5m), then their eyeballs wouldn't need to be on the ground, and the estimated height would be 720m. :)
 
I agree with the cat. Probably a typo, somewhere.

I worked it bit differently (\(\displaystyle x = \overline{DE}\) ) but also got 652m, rounded.

Did you notice the surveyor taking "line of sight" has their eyeballs on the ground? Were they to move closer to the vertical stick (4.41m away, instead of 5m), then their eyeballs wouldn't need to be on the ground, and the estimated height would be 720m. :)
Good point. Or if we knew the (eye) height of the surveyor, it would be more realistic.
 

Jomo

New member
I agree with the cat. Probably a typo, somewhere.

I worked it bit differently (\(\displaystyle x = \overline{DE}\) ) but also got 652m, rounded.

Did you notice the surveyor taking "line of sight" has their eyeballs on the ground? Were they to move closer to the vertical stick (4.41m away, instead of 5m), then their eyeballs wouldn't need to be on the ground, and the estimated height would be 720m. :)
Otis, I am missing something that you are saying. If the surveyor moved closer to the stick you claim that their eye balls would not have to be on the ground. That confuses me. For example if their eyeball was 1m above the ground, then wouldn't this yield a different result from their eyeballs being 1.1m above the ground? Please enlighten me.
 

Otis

Member
… if their eyeball was 1m above the ground, then wouldn't this yield a different result from their [eyeball] being 1.1m above the ground?
Yes, changing the eyeball height will yield different results, but I didn't state a specific eyeball height. I just moved the point on the ground (to obtain a rounded answer of 720m). The second surveyor could then be placed somewhere inbetween the new ground point and the stick. While changing the exercise to match the answer, I wanted the second surveyor's eyes located above ground. Otherwise, the poor chap would need to sweat, digging a big hole in the thawing tundra.

However, I did mispeak about the new location of that surveyor. My adjustment moved the point on the ground closer to the stick, and then I forgot to add the new location of the surveyor to my post. That goof placed the surveyor at the new ground point, instead.

When I realized my sloppiness, I tried to edit my post, but I'd been logged out. My second mistake was to fix a mug of tea, first. Just after pouring the hot water, the power went out! (Snowstorm 3 of 3 was in progress.) I waited almost two hours for power to return, but eventually I gave up and went to bed. (It's so nice, waking up to electricity and heat.)

Okay, here's what I could have posted:

Two surveyors estimate the height of a nearby hill. The first surveyor holds a 3m stick on the ground, vertically. The second surveyor moves farther away from the hill, standing 1m away from the stick. Upon taking a line-of-sight reading to the top of the hill (using a laser beam), the second surveyor notes two measurements: (1) the line-of-sight laser beam hits the ground 4.4m away from the stick, and (2) the line-of-sight laser beam passes the vertical stick at 2.4m. A third measurement gives the distance from the top of the stick to the top of the hill as 1498.3m. How high is their estimate of the hill (rounded to the nearest metre)?

Bonus question for you, Jomo. Using my new version, can you say whether the second surveyor's eyeballs are (a) more than 6ft above ground or (b) less than 6ft above ground? :)
 

Otis

Member
… The [scenario] simplifies to a quadratic equation … so it could be solved by a Year 10 student (using the quadratic formula).
Yes, I had assumed that students are familiar with the Quadratic Formula and that scientific calculators are allowed for doing arithmetic and square roots with decimal numbers.
 
Yes, changing the eyeball height will yield different results, but I didn't state a specific eyeball height. I just moved the point on the ground (to obtain a rounded answer of 720m). The second surveyor could then be placed somewhere inbetween the new ground point and the stick. While changing the exercise to match the answer, I wanted the second surveyor's eyes located above ground. Otherwise, the poor chap would need to sweat, digging a big hole in the thawing tundra.

However, I did mispeak about the new location of that surveyor. My adjustment moved the point on the ground closer to the stick, and then I forgot to add the new location of the surveyor to my post. That goof placed the surveyor at the new ground point, instead.

When I realized my sloppiness, I tried to edit my post, but I'd been logged out. My second mistake was to fix a mug of tea, first. Just after pouring the hot water, the power went out! (Snowstorm 3 of 3 was in progress.) I waited almost two hours for power to return, but eventually I gave up and went to bed. (It's so nice, waking up to electricity and heat.)

Okay, here's what I could have posted:

Two surveyors estimate the height of a nearby hill. The first surveyor holds a 3m stick on the ground, vertically. The second surveyor moves farther away from the hill, standing 1m away from the stick. Upon taking a line-of-sight reading to the top of the hill (using a laser beam), the second surveyor notes two measurements: (1) the line-of-sight laser beam hits the ground 4.4m away from the stick, and (2) the line-of-sight laser beam passes the vertical stick at 2.4m. A third measurement gives the distance from the top of the stick to the top of the hill as 1498.3m. How high is their estimate of the hill (rounded to the nearest metre)?

Bonus question for you, Jomo. Using my new version, can you say whether the second surveyor's eyeballs are (a) more than 6ft above ground or (b) less than 6ft above ground? :)
Did you get to drink your tea in the dark?
 

Otis

Member
Did you get to drink your tea in the dark?
Yes, thank you very much. I drank it somewhat quickly, while pondering how much time I had before the cubbyhole in the attic space (where I sleep) would become refridgerated. I got into bed with a small bottle of Scotch.

Today, we have 14 inches of snow on the ground, and it's raining.
 
Last edited:

Masaru

New member
Thank you, but not entirely satisfied...

Unless I am making the same mistake as you, I agree with your answer of 652m. (Textbooks have been known to be wrong!)

The equation simplifies to a quadratic equation when you square both sides so it could be solved by a Year 10 student (using the quadratic formula).
Thank you for letting me know that your answer is also the same as mine.

But I must let you know that this is a question designed for Year 10 students in NEW ZEALAND!

It is not for US or UK or even Australia where they learn things faster and do more advanced learning than in New Zealand.

In New Zealand, at Year 10, you NEVER ever learn quadratic formula - it is acutally for Year 12 students here in New Zealand.

Therefore, it is beyond my comprehension that they put this kind of question in Year 10 New Zealand textbook.

On top of that, as you pointed out, the answer provided at the back of the text seems to be wrong...

Pathetic, indeed.
 

Dr.Peterson

Active member
Thank you for letting me know that your answer is also the same as mine.

But I must let you know that this is a question designed for Year 10 students in NEW ZEALAND!

It is not for US or UK or even Australia where they learn things faster and do more advanced learning than in New Zealand.

In New Zealand, at Year 10, you NEVER ever learn quadratic formula - it is acutally for Year 12 students here in New Zealand.

Therefore, it is beyond my comprehension that they put this kind of question in Year 10 New Zealand textbook.

On top of that, as you pointed out, the answer provided at the back of the text seems to be wrong...

Pathetic, indeed.
Working backward from the answer, I see that if the hill is 720 meters high, then 1500 m is the distance from the eye to the hill, horizontally.

It's possible that the question itself was somehow mangled, and was meant to be far simpler; or that whoever wrote the answer misread it, making it easier.

In any case, the publisher's proofreading staff are not doing their job.
 

Otis

Member
… it is beyond my comprehension that they put this kind of question in Year 10 New Zealand textbook …
I thought of a different approach, and it doesn't require solving a quadratic equation.

Because 0.6 is very small compared to 1500, we could estimate the distance from the 2.4m mark (on the stick) to the top of the hill as 1500m. This shortcut would introduce error, but we're rounding an estimated result, either way.

Using your diagram's labels:

Calculate \(\displaystyle \overline{AC}\) by Pythagorean Theorem

Therefore, \(\displaystyle \overline{AF}\) = \(\displaystyle \overline{AC}\) + 1500

Solve the proportion:

\(\displaystyle \dfrac{\overline{BC}}{\overline{AC}} = \dfrac{x}{\overline{AF}}\)


Whether I carry four decimal places through intermediate calculations or use exact values throughout, my final result (≈651.49660) rounds to 651m.

Of course, the book's answer is still wrong, but at least New Zealanders in year 10 have a way to simplify and solve the exercise (±0.5m). :cool:
 
Yes, thank you very much. I drank it somewhat quickly, while pondering how much time I had before the cubbyhole in the attic space (where I sleep) would become refridgerated. I got into bed with a small bottle of Scotch.

Today, we have 14 inches of snow on the ground, and it's raining.
Temperature here was 34 degrees C (93 degrees F) yesterday and that's nowhere near our hottest day.
Would love some snow! (The grass is always greener on the other side of the fence, hey?!)
 

Masaru

New member
Thank you for your deep insight.

Working backward from the answer, I see that if the hill is 720 meters high, then 1500 m is the distance from the eye to the hill, horizontally.

It's possible that the question itself was somehow mangled, and was meant to be far simpler; or that whoever wrote the answer misread it, making it easier.

In any case, the publisher's proofreading staff are not doing their job.
Oh, I did not notice that the answer would be, as you say, 720 meters, if the distance of 1500 m referred to the horizontal distance from the eye to the hill.

However, it is so clear from the question that this distance is not the horizontal one because it says:
"The distance from the top of the stick to the top of the hill is 1500 m".

So I totally agree with you when you say that the
publisher's proofreading staff are not doing their job.
 

Masaru

New member
Thank you for your help.

I thought of a different approach, and it doesn't require solving a quadratic equation.

Because 0.6 is very small compared to 1500, we could estimate the distance from the 2.4m mark (on the stick) to the top of the hill as 1500m. This shortcut would introduce error, but we're rounding an estimated result, either way.

Using your diagram's labels:

Calculate \(\displaystyle \overline{AC}\) by Pythagorean Theorem

Therefore, \(\displaystyle \overline{AF}\) = \(\displaystyle \overline{AC}\) + 1500

Solve the proportion:

\(\displaystyle \dfrac{\overline{BC}}{\overline{AC}} = \dfrac{x}{\overline{AF}}\)


Whether I carry four decimal places through intermediate calculations or use exact values throughout, my final result (≈651.49660) rounds to 651m.

Of course, the book's answer is still wrong, but at least New Zealanders in year 10 have a way to simplify and solve the exercise (±0.5m). :cool:
Your method should be perfectly appropriate for this question because it says "How high, correct to the nearest metre, is their ESTIMATE of the height of the hill?"

So we are not expected to give a precise answer but it has to be an estimate.

Having said that, our answer of 651 m is still not the answer provided in the text, which is so confusing.
 

Masaru

New member
The name of this textbook that you should not use!

Just interested....what's the textbook?
The name of the textbook is:
New Zealand Mathematics second edition 10

Published by Haese & Harris Publications Ltd (Auckland New Zealand)

This is a second edition published long time ago back in 2002, so the current one might be better with corrections, but I am not sure...

The thing is that the proofreaders did not do their job properly!

(By the way, here in New Zealand, most of textbooks are so terrible with full of errors. Quite a long time ago, when I was working on a Year 9 New Zealand mathematics textbook, I found about 50 errors even if I had not finished 2/3 of the book! So I rang up the publisher to complain about it, but they have no manner whatsoever so they did not even apologize for so many errors in the book. They also told me that they would make corrections. However, even though that was about a decade ago, and there are still many errors in the books with no correction whatsoever. The customer service in New Zealand is so notorious and bad enough for me to escape from this country for good next month.)
 

Dr.Peterson

Active member
Your method should be perfectly appropriate for this question because it says "How high, correct to the nearest metre, is their ESTIMATE of the height of the hill?"

So we are not expected to give a precise answer but it has to be an estimate.

Having said that, our answer of 651 m is still not the answer provided in the text, which is so confusing.
That brings up another really odd thing about the problem.

It says, "The distance from the top of the stick to the top of the hill is 1500 m (as measured by laser equipment). How height, correct to the nearest metre, is their estimate of the height of the hill?"

If they have laser measurements of a distance, why would surveyors be satisfied with such a rough estimate of height, and such a cheap way to measure angles (holding up a stick, rather than using a transit)?

Textbooks in general are not known for real-life problems, but this is particularly odd.

More:

I located the book here: http://paknet1.pakuranga.school.nz/matdocs/nz10book.pdf, and see that you quoted everything correctly (page 136, Exercise 5A #9). I also found the same question on another site from a few years ago, giving 652 as the answer, presumably from a later edition.

Finally, the problem is much easier if you suppose that "the top of the stick" means the 2.4 meter point on the line of sight. Then it's just Pythagoras and similar triangles (the subject that was just covered). And knowing the context, we would have been less likely to look for advanced methods and difficult interpretations. (That doesn't excuse bad wording and wrong answers.)
 

Masaru

New member
This is a very odd unrealistic question as you say.

That brings up another really odd thing about the problem.

It says, "The distance from the top of the stick to the top of the hill is 1500 m (as measured by laser equipment). How height, correct to the nearest metre, is their estimate of the height of the hill?"

If they have laser measurements of a distance, why would surveyors be satisfied with such a rough estimate of height, and such a cheap way to measure angles (holding up a stick, rather than using a transit)?

Textbooks in general are not known for real-life problems, but this is particularly odd.

More:

I located the book here: http://paknet1.pakuranga.school.nz/matdocs/nz10book.pdf, and see that you quoted everything correctly (page 136, Exercise 5A #9). I also found the same question on another site from a few years ago, giving 652 as the answer, presumably from a later edition.

Finally, the problem is much easier if you suppose that "the top of the stick" means the 2.4 meter point on the line of sight. Then it's just Pythagoras and similar triangles (the subject that was just covered). And knowing the context, we would have been less likely to look for advanced methods and difficult interpretations. (That doesn't excuse bad wording and wrong answers.)
Yes, I agree with you.

I wonder why anyone would use this kind of backward method using a stick to measure the height of the hill when they are acually using laser to accurately measure the distance from the stick to the top of the hill?

It just does not make sense to me because they could use only laser to measure the height of the hill - no need of using a stick!

You also mentioned that you
also had found the same question on another site from a few years ago, giving 652 as the answer, presumably from a later edition, which indicates that our initial answer is correct and they have made correction in their later edition.

Thank you so much for your effort to locate this information which proves that our answer is not wrong, and what is wrong is the wording and incorrect answer in the textbook!

 

mmm4444bot

Super Moderator
… I wonder why anyone would use this kind of backward method using a stick to measure the height of the hill when they are acually using laser to accurately measure the distance from the stick to the top of the hill? …
Some story problems are manufactured; common-sense isn't always part of the agenda. ;)
 
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