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Trig Problem(Law of Sines)

  1. Nov 28, 2003 #1
    2 boys desiring to estimate the height of a nearby radio tower measured the angle of elevation at their house and found it to be 52 degrees. they took a second measurement from the second-story window and found the angle of elevation to be 44 degrees. they next measured the window to be 24 feet above the ground. to the nearest foot, what is the height of the radio tower????
    Law of sines:::

    i cant figure out how to use new latex for this ........
    sin a / a = sin b / b = sin c / c

    thanks
     
  2. jcsd
  3. Nov 29, 2003 #2

    ShawnD

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    A diagram of what you mean would really really help.
     
  4. Nov 29, 2003 #3

    AD

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    I'm pathetic with Latex as well, so bear with me.

    2 boys desiring to estimate the height of a nearby radio tower measured the angle of elevation at their house and found it to be 52 degrees. they took a second measurement from the second-story window and found the angle of elevation to be 44 degrees. they next measured the window to be 24 feet above the ground. to the nearest foot, what is the height of the radio tower?

    Okay, call the height of the tower 'a' and the distance along the ground toward the tower 'c'. You need to think of a right-angled triangle with base 'c' and opposite leg 'a'.

    When the boys take their measurements from the second-floor window, think of a different triangle, this one with opposite side (a - 24)ft in height. You may notice that the base 'c' does not change between these two locations.

    Therefore

    a/(sin 52) = c/(sin (90 - 52)) = c/(sin 38)

    and

    (a - 24)/(sin 44) = c/(sin(90 - 44)) = c/(sin 46)

    Since

    a/(sin 52) = c/(sin 38)

    a = c(sin 52)/(sin 38)

    And also, since

    (a - 24)/(sin 44) = c/(sin 46)

    a = c(sin 44)/(sin 46) + 24

    So, now we know that

    c(sin 44)/(sin 46) + 24 = c(sin 52)/(sin 38)

    And that

    c = 24/[(sin 52)/(sin 38) - (sin 44)/(sin 46)]

    From before, we found that

    a = c(sin 52)/(sin 38)

    So substitute your value for c into this equation and that will yield your answer.
     
  5. Nov 29, 2003 #4
    AD- thanks for the calculations. but i still dont understand how to diagram the second triangle? the first triangle is a right triangle,
    i think i have the first triangle right..but where does the 2nd one go?
     
  6. Nov 29, 2003 #5

    AD

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    I've drawn a diagram in paint, but it is not attaching because it is apparently too big, despite the fact that I've gotten the file size down to 9KB. Perhaps you could PM me your e-mail address and I can send it to you that way.
     
  7. Nov 29, 2003 #6

    AD

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    Did you get the e-mail, Starky? Or did it not go through?
     
  8. Nov 29, 2003 #7
    oh ok! thanks so much for the graph, now i understand what your talking about with that diagram. i can't believe i couldn't figure that out- it's so easy now that i look at your picture- thanks again andrew.
    sincerely,
    ~david
     
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