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    Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

    Thanks
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    (Original post by ThrashMetal)
    Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

    Thanks
    Posted from TSR Mobile
    \angle BOC = \pi - \theta

    So, there's where pi comes in.

    Post your initial equation, and we can check.
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    (Original post by ThrashMetal)
    Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

    Thanks
    Posted from TSR Mobile
    Well, usually, angles on a straight line add to 180, but 180 degrees is \pi radians.

    You know the radius and the relationship between the two sectors (in terms of area)

    Does that help?
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    I think I've done it. I forgot half a circle was pi radians lol.

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    (Original post by ThrashMetal)
    I think I've done it. I forgot half a circle was pi radians lol.

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    Looking good

    A bit of a mess on the page, though
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    (Original post by Indeterminate)
    Looking good

    A bit of a mess on the page, though
    Haha yeah I always answer the question roughly then write it out neatly
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    (Original post by ThrashMetal)
    Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

    Thanks
    Posted from TSR Mobile
    AB is the diameter of the circle and so bisects the circle (perhaps the point you are missing).
    The area of the whole circle is pi x r^2 = pi x 4^2 = 16pi
    Half the area of the whole circle is (1/2) x 16pi = 8pi
    So we have (Area of AOC Sector) + (Area of BOC Sector) = 8pi
    But (Area of AOC Sector) + 5 = (Area of BOC Sector)
    So (Area of AOC Sector) + ( (Area of AOC Sector) + 5) = 8 pi
    So 2 x (Area of AOC Sector) = 8pi - 5.
    But: (Area of AOC Sector) = (1/2) (r^2)(theta)=(1/2)(4^2)(theta)= 8 x (theta)

    So we have 2 x (8 x (theta)) = 16 (theta) = 8pi -5
    Thus (theta) = (8pi - 5)/16.

    Hope this helps
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    Thanks for the help guys. Got there in the end. Plus repped
 
 
 
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