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Math Help - URGENT HELP! cirlcess

  1. #1
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    URGENT HELP! cirlcess

    im not sure how i can draw the specific image i have for the problem to be solved, but i'll try to explain it and then maybe someone can help me out with this. so there is a circle with radius r and center O. The points A, B, and C are on the circle and AOC = the angle θ, situated across the arc ABC. The area of the sector OABC is 4/3π and the length of the arc ABC is 2/3π.

    Find the value of r and θ.

    Thanks for your help!
    Last edited by miley_22; October 13th 2007 at 07:51 AM. Reason: made a mistake in typing values
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  2. #2
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    The area of the sector is \frac{1}{2}r^{2}{\theta}=\frac{4\pi}{3}

    The length of the curve is: r{\theta}=\frac{2\pi}{3}

    Using the two equations, solve for r and theta.

    BTW, use proper grouping symbols. The way you have them written would suggest \frac{4}{{\pi}3}. I assume you mean what I have above?.
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  3. #3
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    Quote Originally Posted by miley_22 View Post
    im not sure how i can draw the specific image i have for the problem to be solved, but i'll try to explain it and then maybe someone can help me out with this. so there is a circle with radius r and center O. The points A, B, and C are on the circle and AOC = the angle θ, situated across the arc ABC. The area of the sector OABC is 4/3π and the length of the arc ABC is 2/3π.

    Find the value of r and θ.

    Thanks for your help!
    Area of Sector OABC = [(2pi -theta)/(2pi)]*pi(r^2) = (4/3)pi
    [(2pi)/(2pi) -theta/(2pi)]*pi(r^2) = 4pi/3
    [1 -theta/(2pi)](r^2) = 4/3 -----------------(1)

    Arc ABC = r(theta) = (2/3)pi
    theta = 2pi/(3r) -----------------------(2)
    Substitute that into (1),
    [1 -(2pi)/(3r)/(2pi)]r^2 = 4/3
    [1 -1/(3r)]r^2 = 4/3
    r^2 -r/3 = 4/3
    Clear the fractions, multiply both sides by 3,
    3r^2 -r = 4
    3r^2 -r -4 = 0
    Factor that, ------- or use the Quadratic Formula
    (3r -4)(r +1) = 0
    r = 4/3 or -1

    Reject r = -1 because there are no negative dimensions.

    So, r = 4/3 units -------------------answer.

    theta = 2pi/(3r) -------------(2)
    theta = 2pi/[3(4/3)]
    theta = pi/2 radians -------------------answer.
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  4. #4
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    Hello, miley_22!

    Another approach . . .


    There is a circle with radius r and center O.
    The points A, B, and C are on the circle
    and \theta = \angle  AOC subtends arc \widehat{ABC}.

    The area of the sector \overline{OABC}\:=\:\frac{4}{3}\pi
    . . and the length of the arc \widehat{ABC} \:=\:\frac{2}{3}\pi

    Find the value of r and \theta.
    Code:
                  * * *
              *           *   A
            *               o
           *              /  o B
                        /
          *           / θ     *
          *         * - - - - o C
          *         O         *
    
           *                 *
            *               *
              *           *
                  * * *

    The area of the sector is: . A \:=\:\frac{1}{2}r^2\theta \:=\:\frac{4}{3}\pi .[1]

    The arc length of \widehat{ABC} is: . s \:=\:r\theta\:=\:\frac{2\pi}{3} .[2]

    Divide [1] by [2]: . \frac{\frac{1}{2}r^2\theta}{r\theta} \:=\:\frac{\frac{4\pi}{3}}{\frac{2\pi}{3}}\quad\Ri  ghtarrow\quad\frac{1}{2}r\:=\:2\quad\Rightarrow\qu  ad\boxed{ r \:=\:4}

    Substitute into [2]: . 4\theta\:=\:\frac{2\pi}{3}\quad\Rightarrow\boxed{\  theta \:=\:\frac{\pi}{6}}

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