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Trigonometry - Hot Air Balloon - Homework Question

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Hello all, I am having trouble with the following homework question using trigonometric formulas and the Sine Law.

    Here is the problem:

    Jack spotted a hot air balloon from his house at angle of elevation 57 degrees. Jack's friend who lives 26 km from Jack's house spotted the same balloon at angle of elevation 83 degrees. Determine the altitude of the balloon to the nearest tenth.


    2. Relevant equations
    Sine Laws: [(sinA/a)=(sinB/b)=(sinC/c)], Trigonometric Functions: (SOH, CAH, TOA)


    3. The attempt at a solution

    Here is my solution:


    [​IMG]


    However, the answer given in the text was 49.4 km.
    Whereas, my answer is 33.7 km.

    Can anybody tell me where I went wrong.

    Thanks a lot.
     
  2. jcsd
  3. Aug 10, 2008 #2
    Hmm...I get the same answer you did. I used vectors to set up the equations like this:

    The horizontal components of side a and c should add up to 26.

    c*cos(83) + a*cos(57) = 26

    The vertical components of a and c should be equal.

    h = c*sin(83) = a*sin(57)

    You can solve those two equations for a and c. I get for a the same answer you did. This leads to the same height that you got.

    I also considered that maybe the two observers were on the same side of the balloon. The observer at 57 degrees would be further than the observer at 83. However, you get a negative side length which means a solution is not possible in that configuration.
     
  4. Aug 10, 2008 #3
    Yes, I thought so myself.

    Did anyone else get the same answer?
     
  5. Aug 10, 2008 #4

    tiny-tim

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    Hi Sabellic!:smile:

    erm … you've read into the question something that isn't there.

    the elevations are in the same direction, and you've drawn them opposite. :frown:

    Draw a leany-over triangle, and you'll find you get 49.4. :wink:
     
  6. Aug 10, 2008 #5
    What's a leany-over triangle? When I think of leany-over triangle I think of an obtuse triangle because the point extends laterally over the base as if it is "leaning". But all of the angles are less than 90 degrees.
     
  7. Aug 10, 2008 #6

    tiny-tim

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    Yes … but one of them is an external angle! :smile:

    (Where in the question does it say that they're facing in opposite direcitons? :rolleyes:)
     
  8. Aug 10, 2008 #7
    Doh. Yep that works. I considered that case but got a negative range for one of the sides the first time. Just did it again and it works out.
     
  9. Aug 10, 2008 #8
    Okay, I got the answer. Thanks.

    However, if I didn't know the answer I would have thought the opposite directions is perfectly plausible. They didn't specify same directions. They didn't specify opposite directions.

    What would be the right method of finding the solution? They didn't specify that one of them is an external angle. Should I solve the question two different ways and then see which one is best?
     
  10. Aug 10, 2008 #9

    tiny-tim

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    Well, they didn't specify that it was one-dimensional either, did they?

    If the question uses a word like "elevation" twice without saying what it's an elevation from, then you should assume that it's from the same thing each time … eg north and north, not north and south, or north and east. :smile:
     
  11. Aug 10, 2008 #10
    Very good advice. Thanks.
     
  12. Aug 10, 2008 #11
    Um, I'm getting height = 33.7. Basically I let x be the distance from the 57 deg angle to the foot of the perpendicular and 26-x be the distance from the 83 deg angle to the foot.

    Then it's just a system of two equations with two variables involving tan.
     
  13. Aug 10, 2008 #12
    If x is the distance from the 57 angle, then 26-x would be the distance to the 83 angle if the observers were on opposite sides like the original poster has in his drawing. If they are on the same side, then the distance becomes x-26.

    The horizontal distance from the guy further out from the balloon is a*cos(57), were a is the slant range to the balloon. The horizontal distance from the guy closer to the balloon is c*cos*(83). The two observers are separated by 26, on the same side, so this gives you

    a*cos(57) – c*cos(83) = 26.

    The vertical distance or height is the same for each, or
    h = a*sin(57) = c*sin(83)

    Solve for c.
    c=a*sin(57)/sin(83)

    Plug in to first equation.
    a*cos(57) – a*sin(57)*cos(83) / sin(83) = 26.

    Solve for a.
    a = 26*sin(83) / (cos(57)*sin(83) – sin(57)*cos(83) ) = 58.86

    So the height is
    a*sin(57) = 49.37
     
  14. Aug 10, 2008 #13

    tiny-tim

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    Hi Chrisas! :smile:

    Nice solution …

    but try to give hints rather than full solutions :smile:

    (you did the same thing with a series summation thread …)
     
  15. Aug 10, 2008 #14
    Dang, I made a wrong assumption. Hmm I remember doing problems involving both cases but don't remember how the wording was different. Oh well same two variable two equation problem pretty much. Though it might be a bit prettier if you applied the law of sines but I rarely did that in precalc for this type of problem :P.
     
  16. Aug 10, 2008 #15
    [​IMG]
    here's how it should look
     
  17. Aug 11, 2008 #16

    tiny-tim

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    Hi Sabellic! :smile:

    Yes, that's it! :smile:

    The only suggestion I'd make is that it's slightly neater if you do all the calculating in one go …

    you aren't asked for a, so just prove that h = 26 sin83º sin57º/sin26º, and then calculate that directly.

    (it has the extra advantage that if you get the wrong result, it's easier to see how to adjust it! :wink:)
     
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