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Math Help - Trig. segmet area+perimiter

  1. #1
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    Trig. segmet area+perimiter

    Hi again, i'm currently studying for my pure maths intermediate level exam and i need some help with a trigonometry question please:

    -------------


    The points A and B lie on the circumference of a circle center O and radius

    r cm, such that the angle AOB = θ
    radians. The sector AOB has area 9cm^2 and the total perimeter of sector AOB is 15cm.

    a)Show that r satisfies the equation 2r^2 - 15r + 18 = o
    b)Find the value of
    θ,explaining why there is only one possible value.

    ------------------
    I know that area of sector = (1/2).(r^2).(θ
    )
    not sure if perimiter of sector would be length of arc
    + radius x2

    Any help is appreciated,thanks!






    *sry i meant sector not segment in thread title
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  2. #2
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    Quote Originally Posted by floyd View Post
    Hi again, i'm currently studying for my pure maths intermediate level exam and i need some help with a trigonometry question please:

    -------------


    The points A and B lie on the circumference of a circle center O and radius

    r cm, such that the angle AOB = θ
    radians. The sector AOB has area 9cm^2 and the total perimeter of sector AOB is 15cm.

    a)Show that r satisfies the equation 2r^2 - 15r + 18 = o
    b)Find the value of
    θ,explaining why there is only one possible value.

    ------------------
    I know that area of sector = (1/2).(r^2).(θ
    )
    not sure if perimiter of sector would be length of arc
    + radius x2

    Any help is appreciated,thanks!

    Yes, all looks good.

    Remember too that the length of the arc is

    \frac{\theta}{2\pi}\cdot 2\pi r = \theta r.


    So the perimeter will be \theta r + 2r = r(\theta + 2).
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Yes, all looks good.

    Remember too that the length of the arc is

    \frac{\theta}{2\pi}\cdot 2\pi r = \theta r.


    So the perimeter will be \theta r + 2r = r(\theta + 2).
    ok thanks, any idea how i can relate those two equations to solve a) and b)?

    I factorised 2r^2 - 15r + 18 = o which gives r=6,r=3/2 but i still have no idea how to continue
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  4. #4
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    Quote Originally Posted by floyd View Post
    ok thanks, any idea how i can relate those two equations to solve a) and b)?

    I factorised 2r^2 - 15r + 18 = o which gives r=6,r=3/2 but i still have no idea how to continue
    You know the area and perimeter, so solve the equations simultaneously for r and \theta.

    Then substitute r into the equation and see if it holds true.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    You know the area and perimeter, so solve the equations simultaneously for r and \theta.

    Then substitute r into the equation and see if it holds true.
    Ok got it! This is what i did just in case some1 has same problem:

    the two equations are
    1)15 = r(θ+2)
    2)9 = 1/2r^2θ

    from 1) i get θ = 15/r -2 and i substitute it in 2)
    which gives 9= 1/2r^2(15/r - 2)
    9= 1/2(15r - 2r^2)
    18= 15r - 2r^2
    2r^2 - 15r + 18 = o


    part b) by factorising u get r = 6 r = 3/2

    using 1) 15= 6(θ+2) 2) 15= 3/2(θ+2)
    θ = 1/2 θ = 8

    ------
    Thanks,Prove it
    The last thing I want to ask regarding question b) is why is there only one possible value of θ?
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  6. #6
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    Can you have an angle greater than 2\pi radians?
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Can you have an angle greater than 2\pi radians?
    ah i see, so that means i choose the 1/2. Ok thanks!
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