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 (x2.4)/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.
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 (x2.4)/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.
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