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Math Help - Trig word problem - angle of depression.

  1. #1
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    Trig word problem - angle of depression.

    Here is the word problem.

    A bird on top of a vertical pole looked South at an angle of depression of 35 Degrees, saw Icing, Maya's bunny on the ground. The bird turned and looked East, and at an angle of depression of 40 degrees saw, Notowada, Maya's turtle on the ground. The distance between Icing and Ntwoada is 60m. Find the height of the pole given this information.

    I have diagrammed this out and see it perfectly I just don't know how to calculate this given the information.

    I can't see how i can use Sin Rule, Cos rule or anything...Can someone show me the process of doing this. My diagram looks like a right triangular pyramid.

    Please clearly explain your work so I can understand why you are performing certain operations.

    Thank you so much in advance!

    Sincerely,
    Raymond
    Last edited by mr fantastic; March 29th 2011 at 05:40 AM. Reason: Title.
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  2. #2
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    Quote Originally Posted by raymac62 View Post
    Here is the word problem.

    A bird on top of a vertical pole looked South at an angle of depression of 35 Degrees, saw Icing, Maya's bunny on the ground. The bird turned and looked East, and at an angle of depression of 40 degrees saw, Notowada, Maya's turtle on the ground. The distance between Icing and Ntwoada is 60m. Find the height of the pole given this information.

    I have diagrammed this out and see it perfectly I just don't know how to calculate this given the information.

    I can't see how i can use Sin Rule, Cos rule or anything...Can someone show me the process of doing this. My diagram looks like a right triangular pyramid.

    Please clearly explain your work so I can understand why you are performing certain operations.

    Thank you so much in advance!

    Sincerely,
    Raymond
    let h = pole height , x = distance from base of the pole to bunny , y = distance from base of the pole to turtle

    \tan(35) = \dfrac{h}{x}<br />

    \tan(40) = \dfrac{h}{y}<br />

    x^2 + y^2 = 60^2

    solve the system of equations for h
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  3. #3
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    Quote Originally Posted by skeeter View Post
    let h = pole height , x = distance from base of the pole to bunny , y = distance from base of the pole to turtle

    \tan(35) = \dfrac{h}{x}<br />

    \tan(40) = \dfrac{h}{y}<br />

    x^2 + y^2 = 60^2

    solve the system of equations for h
    Hi Skeeter,

    Thanks for adressing my question. I have never solved systems of equations with 3 variables, let alone in such a complex form. I know we are dealing with a 3 dimension problem here. Could you please demonstrate and explain how to solve the system of equations and solve for H. I have been obsessing over this question all day and I am so perplexed.

    Thanks in advance,

    Ray
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  4. #4
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    \tan(35) = \dfrac{h}{x} \implies x = \dfrac{h}{\tan(35)}

    \tan(40) = \dfrac{h}{y} \implies y = \dfrac{h}{\tan(40)}

    x^2 + y^2 = 60^2

    \left[\dfrac{h}{\tan(35)}\right]^2 + \left[\dfrac{h}{\tan(40)}\right]^2 = 60^2

    \dfrac{h^2}{\tan^2(35)} + \dfrac{h^2}{\tan^2(40)} = 60^2

    h^2 \left[\dfrac{1}{\tan^2(35)} + \dfrac{1}{\tan^2(40)}\right] = 60^2

    h^2[\cot^2(35) + \cot^2(40)] = 60^2

    h^2 = \dfrac{60^2}{\cot^2(35)+\cot^2(40)}

    h = \dfrac{60}{\sqrt{\cot^2(35) + \cot^2(40)}}
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  5. #5
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    Question adjacent/opposite ?? h/x?

    Quote Originally Posted by skeeter View Post
    \tan(35) = \dfrac{h}{x} \implies x = \dfrac{h}{\tan(35)}

    \tan(40) = \dfrac{h}{y} \implies y = \dfrac{h}{\tan(40)}

    x^2 + y^2 = 60^2

    \left[\dfrac{h}{\tan(35)}\right]^2 + \left[\dfrac{h}{\tan(40)}\right]^2 = 60^2

    \dfrac{h^2}{\tan^2(35)} + \dfrac{h^2}{\tan^2(40)} = 60^2

    h^2 \left[\dfrac{1}{\tan^2(35)} + \dfrac{1}{\tan^2(40)}\right] = 60^2

    h^2[\cot^2(35) + \cot^2(40)] = 60^2

    h^2 = \dfrac{60^2}{\cot^2(35)+\cot^2(40)}

    h = \dfrac{60}{\sqrt{\cot^2(35) + \cot^2(40)}}


    Why: \tan(35) = \dfrac{h}{x} ?

    That's adjacent/opposite. I thought tan A was opposite/adjacent.

    Did you mean to say \tan(55) = \dfrac{h}{x} and \tan(50) = \dfrac{h}{y} ??
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  6. #6
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    Quote Originally Posted by raymac62 View Post
    Why: \tan(35) = \dfrac{h}{x} ?

    That's adjacent/opposite. I thought tan A was opposite/adjacent.

    Did you mean to say \tan(55) = \dfrac{h}{x} and \tan(50) = \dfrac{h}{y} ??
    no, I did not mean that. remember alternate interior angles? the angle of depression = the angle of elevation. look at the sketch ...
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  7. #7
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    I see! Thank you very much.
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