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    I have been stuck on this question for hours...
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    The way I was going to work this out was to find the area of the triangle in figure 1 and then double it and take away the area 'S' seen in figure 4. So,
    \displaystyle A(\pi/3) = 2(0.5*G*1)-'S'

    For the area of the sector, from figure two you can see that the angle is \displaystyle \pi-(\pi/3).
    So \displaystyle  0.5(1*1)(2\pi/3) = \pi/3 = S

    In order to work out the triangle in figure one, I worked out 'G' from figure 2.
    \displaystyle Cos(x)=A/H.The hypotenuse 'OP' has already been worked out from a previous part of the question where the X-Coord of P:
    \displaystyle 

 P= csc(\theta)+cot(\theta)+1.

Cos(\theta)[csc(\theta)+cot(\theta)+1] = G

x=\pi/3



G= (\sqrt{3}+ 1)/2

    Area of the triangle in figure one is g/2. Multiplied by 2 gives the whole ares including the sector (figure 2).

    \displaystyle

So A(\pi/3)=\frac{(\sqrt{3}+1)}{2}- \frac{\pi}{3}
    Which is not the right answer.
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    (Original post by KloppOClock)
    ...
    Sorry, I can't see all the attachments, so I don't know what your working is refering to.

    However.

    Let C be the centre of the circle. And drop a perpendicular from C to OP meeting it at M, say.

    Then, agree area of sector CQM = pi/3

    Area of quadrilateral CQPM is twice area of triangle CPM.

    CPM is a 30,60,90 triangle and since CM=1, it follows MP = root(3)

    Area CPM is thus root(3) / 2, and area of quadrilateral is root(3).

    Hence....
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    (Original post by ghostwalker)
    Sorry, I can't see all the attachments, so I don't know what your working is refering to.

    However.

    Let C be the centre of the circle. And drop a perpendicular from C to OP meeting it at M, say.

    Then, agree area of sector CQM = pi/3

    Area of quadrilateral CQPM is twice area of triangle CPM.

    CPM is a 30,60,90 triangle and since CM=1, it follows MP = root(3)

    Area CPM is thus root(3) / 2, and area of quadrilateral is root(3).

    Hence....
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    (Original post by KloppOClock)
    ...
    See my previous post for a working.

    Regarding your working:

    You've used diagram 3, when working out G, using OP as the hypoteneuse.

    However, diagram 3 is incorrect. The perpendicular to RP at Q, does not go through O. It would only go through O if the angle theta was 45 degrees. There's your problem.
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    (Original post by ghostwalker)
    See my previous post for a working.

    Regarding your working:

    You've used diagram 3, when working out G, using OP as the hypoteneuse.

    However, diagram 3 is incorrect. The perpendicular to RP at Q, does not go through O. It would only go through O if the angle theta was 45 degrees. There's your problem.
    Thank you! I was wondering what I was doing wrong for ages!
 
 
 
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