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Math Help - Trigonometry height and distance problems

  1. #1
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    Question Trigonometry height and distance problems

    A minar stands on a level ground.The shadow of the minar is 30 meter more when angel of elevation of sun (of sun's altitude) is 45^o than the angel of elevation of sun 60^o .Find height of the minar.

    Plz solve this question.
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  2. #2
    Super Member fardeen_gen's Avatar
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    Let the two triangles formed be ABC and ABC'. [both right angled at B]

    ACB = 60 degrees, AC'B = 45 degrees.

    Let AB = h, BC = x (therefore, CC' = 30)
    Now in triangle ABC,

    tan 60 = AB/BC
    (or) tan 60 = h / x
    (or) sqrt(3) = h / x
    (or) x = h / sqrt(3) --------- I


    In triangle ABC',

    tan 45 = AB / AC'
    (or) 1 = h / (AC + CC')
    (or) 1 = h / (x + 30)
    (or) h = x + 30
    (or) h = h/sqrt(3) + 30 --------using I
    (or) h = [30*sqrt(3)]/[sqrt(3) - 1]

    Therefore height of the minar is (30*1.73)/(0.73) m.
    Last edited by fardeen_gen; July 20th 2008 at 05:38 AM. Reason: Missing Parentheses
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  3. #3
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    Thanks for that fardeen_gen
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  4. #4
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    i don't understand

    Quote Originally Posted by fardeen_gen View Post
    Let the two triangles formed be ABC and ABC'. [both right angled at B]

    ACB = 60 degrees, AC'B = 45 degrees.

    Let AB = h, BC = x (therefore, CC' = 30)
    Now in triangle ABC,

    tan 60 = AB/BC
    (or) tan 60 = h / x
    (or) sqrt(3) = h / x
    (or) x = h / sqrt(3) --------- I


    In triangle ABC',

    tan 45 = AB / AC'
    (or) 1 = h / (AC + CC')
    (or) 1 = h / (x + 30)
    (or) h = x + 30
    (or) h = h/sqrt(3) + 30 --------using I
    (or) h = [30*sqrt(3)]/[sqrt(3) - 1]

    Therefore height of the minar is (30*1.73)/(0.73) m.
    hey fardeen, i don't understand the last bit that i have highlighted. why is h=h/sqrt(3) + 30. I don't understand why you have to multiply the 30m by tan60 and why you have to divide that by tan35 - tan45. could you please explain why you did these calculations a.s.a.p please i'm stuck. and how is h=x+30? ty
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