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    Hey.. I'm a bit stuck on this one, don't know what to do:
    In the green heinemann book, review ex 2, p.176, question 47

    There's a diagram which you might need to see to answer the question, but here's the question anyway

    A semicircle, has midpoint O of the diameter AB.
    The point P is on the semicircle such that the area of sector POB is equal to the area enclosed by the segment AP

    The angle POB is x radians
    Prove that x = (1/2)(pi - sin x)

    anyone?
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    Area of shaded segment = area of sector AOP - area of triangle AOP

    = 1/2(pi - x)r^2 - 1/2(r^2)(sin (pi - x))
    = r^2/2 (pi-x - sin (pi-x))

    Area of sector POB = 1/2(x)(r^2)

    r^2/2 (pi-x - sin (pi-x)) = 1/2(x)(r^2)
    pi - x - sin (pi - x) = x
    2x = pi - (sin (pi-x))
    x = 1/2 (pi - sin x)

    I think?
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    (Original post by mik1a)
    Area of shaded segment = area of sector AOP - area of triangle AOP

    = 1/2(pi - x)r^2 - 1/2(r^2)(sin (pi - x))
    = r^2/2 (pi-x - sin (pi-x))

    Area of sector POB = 1/2(x)(r^2)

    r^2/2 (pi-x - sin (pi-x)) = 1/2(x)(r^2)
    pi - x - sin (pi - x) = x
    2x = pi - (sin (pi-x))
    x = 1/2 (pi - sin x)

    I think?
    guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
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    (Original post by TheWolf)
    guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
    I couldn't think of any other way.
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    Aah, of course 1/2 ab sin C

    Cheers!
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    (Original post by TheWolf)
    guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
    I've been told that you're expected to have P1 knowledge...
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    (Original post by kimoni)
    I've been told that you're expected to have P1 knowledge...
    CRAP i cant remember most of the p1 formulas :P
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    (Original post by TheWolf)
    CRAP i cant remember most of the p1 formulas :P
    me neither.. Although I've just learnt the circle ones from doing this question.
    I sincerely doubt we'll be asked any sequences and series questions, which contained those AP and GP forumlae...
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    (Original post by lgs98jonee)
    wot have u done here??have u said that sin(pi-x)=sin x that isnt true is it? or am i missing sumthing simple ??? mik1a have u even done p2 yet? - as in started it at school??
    If you havent done P2 you won't get this...
    sin(pi - x) expands to sin(pi)cos(x) - cos(pi)sin(x)
    and sin(pi) is 0....
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    (Original post by lgs98jonee)
    erm no actually sinpi=1, so that goes to cosx as cospi=0
    sin(pi) is 0, not 1... and cos(pi) is -1, read it through, and it'll make sense! :rolleyes:
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    (Original post by kimoni)
    sin(pi) is 0, not 1... and cos(pi) is -1, read it through, and it'll make sense! :rolleyes:
    ah yes good point..keep think pi radians=90 degrees
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    (Original post by lgs98jonee)
    ah yes good point..keep think pi radians=90 degrees
    heh, just don't make that mistake in the exam!
 
 
 
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