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Math Help - Finding central angle after slicing cone

  1. #1
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    Finding central angle after slicing cone

    A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector?

    I drew a diagram. I thought the central angle was 90 degrees because of the 6,8,10 triangle that included the slant height. The answer says 216 degrees.
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  2. #2
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    Re: Finding central angle after slicing cone

    Hello, benny92000!

    The problem is not stated clearly.


    A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm.
    If this cone is cut along the slanted height to make a sector,
    what is the central angle, in degrees, of the sector?

    The side view of the cone looks like this:
    Code:
                  B
                  *
                 /:\
                / : \
               /  :  \
              /   :   \ 10
             /    :8   \
            /     :     \
           /      :      \
        A *-------*-------* C
          : - 6 - : - 6 - :
    We see that the slant height is 10 cm.
    The circumference of the base is: . 2\pi r = 2\pi(6) = 12\pi\text{ cm}

    Now we cut the cone and lay it flat.
    We have a large sector of a circle of radius 10.
    Code:
                    B
                  * * *
              *           *
            *               *
           *                 *
    
          *         O         *
          *         *         *
          *        / \        *
               10 /   \ 10
           *     /     \     *
            *   /       \   *
            A *           * C
                  * * *
    Length-of-arc Formula: . s \,=\,r\theta . where:. \begin{Bmatrix}{s &=& \text{length of arc} \\ r &=& \text{radius} \\ \theta &=& \text{central angle}\\[-2mm] && \text{in radians} \end{Bmatrix}

    We have: . \begin{Bmatrix}s \,=\,\text{arc}(ABC) \,=\, 12\pi \\ r \:=\: 10\end{Bmatrix}


    Therefore: . 12\pi \,=\,10\theta

    . . . . . . . . . . \theta \:=\:\frac{12\pi}{10}\text{ radians} \:=\:216^o

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