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Math Help - Circle From Included Angle and Chord Length

  1. #1
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    Circle From Included Angle and Chord Length

    I have the direct length between the start and end of a chord, and I have the included angle. How can I find the radius of the circle the arc is on?
    Thanks
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  2. #2
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    Quote Originally Posted by therealdrag0 View Post
    I have the direct length between the start and end of a chord, and I have the included angle. How can I find the radius of the circle the arc is on?
    Thanks
    you can find the radius using the law of cosines.
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  3. #3
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    Hm. I've looked at the page on Wikipedia: c^2 = a^2 + b^2 - 2*a*b*cos(y)

    (because a=b) -> c^2 = 2a^2 - 2*a^2*cos(y)

    but...I only have y (the angle) and c (the chord length), so I'm not sure what to do from here.
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  4. #4
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    Quote Originally Posted by therealdrag0 View Post
    Hm. I've looked at the page on Wikipedia: c^2 = a^2 + b^2 - 2*a*b*cos(y)

    (because a=b) -> c^2 = 2a^2 - 2*a^2*cos(y)

    but...I only have y (the angle) and c (the chord length), so I'm not sure what to do from here.
    factor out a^2 from the right side ...

    c^2 = a^2(2 - 2\cos{y})

    can you solve for a now?
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  5. #5
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    Ah; good move.

    \sqrt{c^2 / (2 - 2\cos{y})} = a

    Thank you for the help.
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  6. #6
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    Hello, therealdrag0!

    I have the length of a chord, and I have the included angle.
    How can I find the radius of the circle?
    Did you follow skeeter's suggestion?

    Code:
                  * * *    A
              *           *
            *           / | *
           *         r /  |  *
                      /   |
          *          /    |   *
          *       O * θ   |d  *
          *          \    |   *
                      \   |
           *         r \  |  *
            *           \ | *
              *           *
                  * * *    B

    You know: . \begin{array}{ccc}d \:=\:\text{length of chord }AB \\ <br />
\theta \:=\: \text{in{c}luded angle }AOB \end{array}


    Law of Cosines: . d^2 \:=\:r^2 + r^2 - 2r^2\cos\theta

    . . d^2 \:=\: 2r^2 - 2r^2\cos\theta \:=\: 2r^2(1 - \cos\theta) \:=\:4r^2\left(\frac{1-\cos\theta}{2}\right) \quad\Rightarrow\quad d^2\:=\:4r^2\sin^2\!\left(\tfrac{\theta}{2}\right)


    Hence, we have: . d \:=\:2r\sin\tfrac{\theta}{2} \quad\Rightarrow\quad r \:=\:\frac{d}{2\sin\frac{\theta}{2}} \quad\Rightarrow\quad r\;=\;\tfrac{1}{2}d\csc\tfrac{\theta}{2}


    There! . . . We just invented a formula for this problem . . .

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