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Find lengths of triange sides using trig calculations

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    See attached image.


    2. Relevant equations
    Sine law
    Cosine law


    3. The attempt at a solution
    I'm really not sure what to do on this one. I know that the angle ABC and ACB will be the same, but after that I'm stumped. It's been a long time since I've taken a math course so I'm not sure what the relevance of the length of AD is.
     

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  3. Mar 8, 2013 #2

    SteamKing

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    Why don't you draw triangle ACD? Can you identify the lengths of any of the sides of this triangle?
     
  4. Mar 8, 2013 #3

    Ray Vickson

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    AD tells you how tall the pyramid is. If you had AD = 0 it would be flat, with no peak at all. If you had AD = 6000m the thing would be almost as high as Mt. Everest.

    You need to look at some other triangles, such as ADB.
     
  5. Mar 8, 2013 #4
    Oh okay, that makes sense. Except I don't have any angles, and all I have is one measurement. How would I proceed?
     
  6. Mar 8, 2013 #5
    Other than ADB (or ADC) which I think is 90°?
     
  7. Mar 8, 2013 #6

    Ray Vickson

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    Actually, you have one measurement, and one "indirect" measurement---something you can easily figure out from the measurements you _do_ have. Just draw a picture, and look carefully at what you have.
     
  8. Mar 8, 2013 #7
    The 90° angle?
     
  9. Mar 8, 2013 #8
    Or DC is 1m?
     
  10. Mar 8, 2013 #9

    Ray Vickson

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    You need to figure out these things for yourself. We are not allowed to do your work for you.
     
  11. Mar 8, 2013 #10

    Mark44

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    Actually, 6000 m. short of the height of Mt. Everest by nearly 3000 m. The reference I found gives Everest's height as 29,035 ft or 8850 m. I remember the published height being 29,028 ft some years ago, but the Indian plate continues to push up the Himalayas, raising Everest in the process.
     
  12. Mar 8, 2013 #11

    Ray Vickson

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    Sorry: I meant 8000m. Nevertheless, a 6000m high umbrella would still be a sight to see.
     
  13. Mar 8, 2013 #12

    SteamKing

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    Is DC 1m long? Lay out the base and see if this is reasonable.
     
  14. Mar 8, 2013 #13
    I have no idea how to figure it out. If the lesson had properly explained it, then I wouldn't be asking for help. I do not have a teacher to ask these questions to so any help to actually help me understand so that I can then figure out the answer on my own is greatly appreciated.
     
  15. Mar 8, 2013 #14
    I'm not sure what you mean by lay out the base? I haven't taken any math courses in 5 years so I'm a little rusty. Thank you for your help
     
  16. Mar 8, 2013 #15

    Ray Vickson

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    The base is a square of sides 2m. The segment AD is one-half of a diagonal. Can you take it from there? That is, can you calculate the length of AD? If you can , then you will have a right triangle with known base and height, so you can calculate the hypotenuse; that will be the length of AB.
     
  17. Mar 8, 2013 #16
    I'm still not following. I know what AD is, it's 0.5m. I know that the triangle ACD is a right angle triangle. I know once I figure out DC then I can use Pythagorean theorem to find AC, which would be the same as AB. Then I can figure out the angles etc. The problem I'm having right now is figuring out DC. That is all I'm stuck on. Everything else is fine, I just need to know how to start.
     
  18. Mar 8, 2013 #17

    Ray Vickson

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    My apologies: meant that BD is one-half a diagonal. Of course AD is given.
     
  19. Mar 8, 2013 #18
    Oh! Does that mean, I can find BD by using Pyth. theorem and dividing that by 2? So c^2 = 2^2 + 2^2
    ...
    so BD would equal 1.4 m??
    I think I'm finally understand
     
  20. Mar 8, 2013 #19

    Ray Vickson

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    BD is not equal to 1.4; it is equal to √2, and that is approximately 1.4 (but that is a pretty poor approximation). Don't bother using a decimal approximation at this stage; just keep the length as √2 .

    Note: that type of thing can come back to bite you later; it is important to NOT use such approximations except where you need them, especially in larger, more complicated problems.
     
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