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Math Help - Trig triangle/function (ambiguous cases)

  1. #1
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    Exclamation Trig triangle/function (ambiguous cases)

    Hey guys, I need help with another problem,

    I have posted a picture of the triangle

    "For the triangle ABC you are given angle  \beta = 30 degrees, and side c = 3."

    a. Find side b so that one triangle is formed with side b being the altitude of the triangle.

    b. Find side b so that no triangles are formed.

    c. Find side b so that two distinct triangles are formed.

    d. Find side b so that one triangle is formed but side b is not the altitude of the triangle.

    I really need help with these 4 questions so if possible could you guys include work with the answer? Thanks
    Attached Thumbnails Attached Thumbnails Trig triangle/function (ambiguous cases)-untitled.jpg  
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  2. #2
    Super Member bigwave's Avatar
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    Talking

    "For the triangle ABC you are given angle  \beta = 30 degrees, and side c = 3."

    a. Find side b so that one triangle is formed with side b being the altitude of the triangle.

    for b to become the altitude it will have to perpendicular to the base BC

    so 3sin(30\,^{\circ}) = \frac{3}{2} = b
    recall that sin30\,^{\circ} is \frac{1}{2} so multiply it by 3 and you have \frac{3}{2}

    ambiguous triangles ussually use the law of sines to determine the possiblities.
    Last edited by bigwave; December 1st 2009 at 11:56 AM. Reason: spelling
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  3. #3
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    Could you help me with the other ones as well?
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  4. #4
    Super Member bigwave's Avatar
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    hopefully this will help in understanding
    how the law of sines works with ambiguous triangles
    and too that it will address the answers to you other questions.
    see attached image for reference

    use the law of sines, to find possible values of \beta
    if the value of sin\beta > 1
    there is no solution to this equation and no triangle is possible

    if the value of sin\beta = 1,
    there is one (right) triangle possible

    if the value of sin\beta < 1,
    there are two solutions for \beta

    \beta=sin^{-1}\frac{\mathrm{a}sin\mathrm{b}}{\mathrm{a}}\ and \ \beta = 180,^{\circ} - sin^{-1}\frac{\mathrm{a}sin\mathrm{b}}{\mathrm{a}}

    if both of solutions, are substituted into
    \alpha + \beta + \gamma
    result as a positive value for \gamma,
    then two triangles are possilbe
    otherwise there is just one triangle
    SSA is often an ambiguous case.
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  5. #5
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    arg, I can't figure it out! =(
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  6. #6
    Super Member bigwave's Avatar
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    the only answer I can see to this is

    if b is greater than \frac{3}{2} then you have two triangles

    if b is less than \frac{3}{2} there is no triangle

    the way the question is worded is kinda difficult to understand how they want an answer because the only lenght that can be calculated is if it is just one triangle

    at least that is all I can out of this and am sure they are trying to show how the law of sines is used.

    maybe too by now it was explained
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