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Math Help - Area of a Sector in Radians.

  1. #1
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    Area of a Sector in Radians.

    Hi, I need some help with the following two problems:

    1.) Find an exact value for the fraction of the sector represented by the triangle AOB in the sector AOB in the diagram.



    The answer is supposedly 3√3/2π.

    2). The area of the sector AOB and the triangle AOB are at a ratio of 3:2.

    The angle AOB is in radians.

    Show that 2θ-3sinθ=0.




    I have managed to get:

    3=rθ and 2=rsinθ

    Therefore:

    rθ-3=0 and rsinθ-2=0

    But I'm unsure where to go from there.

    Any help would be appreciated. Thanks very much.
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  2. #2
    Senior Member pacman's Avatar
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    For #1.

    Area triangle AOB = ([(3^1/2)/4]r^2)

    Area sector AOB = (1/2)(r^2)(Angle in radians)


    Area triangle AOB/ Area sector AOB = ([(3^1/2)/4]r^2)/(1/2)(r^2)(pi/3),

    r^2 will cancel and 2 and simplifying

    = 3[(3)^1/2]/2pi
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  3. #3
    MHF Contributor
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    Hello evinyssen

    Welcome to Math Help Forum!
    Quote Originally Posted by evinyssen View Post
    2). The area of the sector AOB and the triangle AOB are at a ratio of 3:2.

    The angle AOB is in radians.

    Show that 2θ-3sinθ=0.




    I have managed to get:

    3=rθ and 2=rsinθ See my comment below.

    Therefore:

    rθ-3=0 and rsinθ-2=0

    But I'm unsure where to go from there.

    Any help would be appreciated. Thanks very much.
    Thanks for showing us your working. You're almost there!

    You're right in saying that the area of the sector is \tfrac12r^2\theta, and that the area of the triangle is \tfrac12r^2\sin\theta. But you shouldn't then say that they are equal to 3 and 2 respectively. They are in the ratio
    3:2,which is a different thing.

    We can always write a ratio as a fraction, so do just that:

     \frac{\tfrac12r^2\theta}{\tfrac12r^2\sin\theta}=\f  rac{3}{2}

    Cancel the fraction on the LHS:

    \frac{\theta}{\sin\theta}=\frac{3}{2}

     \Rightarrow 2\theta = 3\sin\theta

    Grandad
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  4. #4
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    Thanks very much for the welcome and the brilliant help!

    I understand both questions now!
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