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OCR (Not MEI) Core 2 Discussion Thread 17th May 2013 + Jan 13 Paper and MS Watch

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    This is for everybody taking the OCR C2 exam on Friday to discuss anything related to the exam such as 'how do you do Q3 from June 2012?' or 'will algebra come up?' or just how you are feeling about it. Good luck to those taking the exam

    Jan 13 QP here:

    http://moodle.eastnorfolk.ac.uk/Math...ary%202013.pdf

    MS:

    http://moodle.eastnorfolk.ac.uk/Math...%20answers.pdf
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    They better make this an easier paper. I've gone over January's and it was solid. :mad:
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    C1 was an easy paper (in comparison to previous years) so this may be as well.
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    As I was doing past papers, January 2013 did seem to be a bit different and more difficult.
    It will be fine.
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    What were the grade boundaries for Jan 2013? I can't seem to find them anywhere.
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    (Original post by simon105)
    What were the grade boundaries for Jan 2013? I can't seem to find them anywhere.
    Core 2:

    (68 for Full UMS)
    (61 for 90 UMS)
    54 for an A
    47 for a B
    40 for a C
    34 for a D
    28 for an E

    Others here:

    http://www.ocr.org.uk/Images/128198-...nuary-2013.pdf
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    18.42 = 10^2+ 20^2 – 2 x 10 x 20 x cos θ
    cos θ = 0.3998
    θ = 66.40

    This is question 4 on Jan 2008 paper. Struggling to see how to rearrange this? Any help would be appreciated.

    Also, good luck for friday guys!
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    (Original post by breakeven)
    18.42 = 10^2+ 20^2 – 2 x 10 x 20 x cos θ
    cos θ = 0.3998
    θ = 66.40

    This is question 4 on Jan 2008 paper. Struggling to see how to rearrange this? Any help would be appreciated.

    Also, good luck for friday guys!
    Is it part iii?
    What do you want to rearrange?
    18.42 = 10^2+ 20^2 – 2 x 10 x 20 x cos θ ?
    I think it should be 18.42^2 when you use cosine rule.
    18.42^2-10^2-20^2=-2 x 10 x 20 x cos θ
    cosθ=(18.42^2-20^2-10^2)/(-2 x 10 x 20)
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    (Original post by purplemind)
    Is it part iii?
    What do you want to rearrange?
    18.42 = 10^2+ 20^2 – 2 x 10 x 20 x cos θ ?
    I think it should be 18.42^2 when you use cosine rule.
    18.42^2-10^2-20^2=-2 x 10 x 20 x cos θ
    cosθ=(18.42^2-20^2-10^2)/(-2 x 10 x 20)
    Have typed this in but im still not getting the right answer? I'm getting cosθ=0.401759

    Might be my calculator, im not sure
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    (Original post by breakeven)
    Have typed this in but im still not getting the right answer? I'm getting cosθ=0.401759

    Might be my calculator, im not sure
    I don't think you should use the rounded value. Maybe try more s.f from part a?
    Where did you get it from? MS?
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    Hi everyone, i was just doing the June 2012 paper and i got stuck on question 7i and 7ii (i've attached) can someone please explain to me how to get the answer- we haven't been taught acute and obtuse angles etc- i was a bit shocked! please help, i'd appreciate it
    Presentation1.ppt
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    (Original post by mia_hilton)
    Hi everyone, i was just doing the June 2012 paper and i got stuck on question 7i and 7ii (i've attached) can someone please explain to me how to get the answer- we haven't been taught acute and obtuse angles etc- i was a bit shocked! please help, i'd appreciate it
    Presentation1.ppt
    On part i. They write tan(x)=2/5 which is Pythagoras. This implies the triangle is right angled, you can then sketch a triangle and label one of the acute (less than 90 degrees) angles 'a'. The opposite and adjacent can then be labelled from soh-cah-toa (Tan(x)=opp/adj). a^2+b^2=c^2 can then be used to give the hyp as root29. Cos(x)=adj/hyp shows cos (x) is root29/5

    working on part ii.

    EDIT: no idea on part ii, sorry. hope someone can help!
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    (Original post by BenChard)
    On part i. They write tan(x)=2/5 which is Pythagoras. This implies the triangle is right angled, you can then sketch a triangle and label one of the acute (less than 90 degrees) angles 'a'. The opposite and adjacent can then be labelled from soh-cah-toa (Tan(x)=opp/adj). a^2+b^2=c^2 can then be used to give the hyp as root29. Cos(x)=adj/hyp shows cos (x) is root29/5

    working on part ii.

    EDIT: no idea on part ii, sorry. hope someone can help!
    Thanks for part i- I get it now ! part ii- well that might require a bit more brain racking on my part thanks and good luck
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    (Original post by mia_hilton)
    Thanks for part i- I get it now ! part ii- well that might require a bit more brain racking on my part thanks and good luck
    No problem, I have just asked some friends who will hopefully know how i'll let you know if i figure it out. You too
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    (Original post by BenChard)
    No problem, I have just asked some friends who will hopefully know how i'll let you know if i figure it out. You too
    Hey- i think i may have figured it out - sin b =3/7 which then means that opp is 3 and hyp is 7 so adjacent is root 40, so cos will be root40/7 (only in the markscheme it says that its -root40/7?) maybe thats got something to do with it being an obtuse angle (which i dont understand)
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    (Original post by mia_hilton)
    Hey- i think i may have figured it out - sin b =3/7 which then means that opp is 3 and hyp is 7 so adjacent is root 40, so cos will be root40/7 (only in the markscheme it says that its -root40/7?) maybe thats got something to do with it being an obtuse angle (which i dont understand)
    If angle is obtuse, it is more than 90 and less than 180.
    From CAST diagram you know, that in the 2nd quarter (more than 90, less than 180, so that's an obtuse angle) sin is positive, cos is negative and tan is negative. This should help.
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    (Original post by mia_hilton)
    Hey- i think i may have figured it out - sin b =3/7 which then means that opp is 3 and hyp is 7 so adjacent is root 40, so cos will be root40/7 (only in the markscheme it says that its -root40/7?) maybe thats got something to do with it being an obtuse angle (which i dont understand)

    (Original post by purplemind)
    If angle is obtuse, it is more than 90 and less than 180.
    From CAST diagram you know, that in the 2nd quarter (more than 90, less than 180, so that's an obtuse angle) sin is positive, cos is negative and tan is negative. This should help.
    I still have no idea here as the angles in a triangle must add up to 180 degrees. right angle=90, obtuse=90+ therefore the other angle is non-existent so we cant use pythag?
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    (Original post by BenChard)
    I still have no idea here as the angles in a triangle must add up to 180 degrees. right angle=90, obtuse=90+ therefore the other angle is non-existent so we cant use pythag?
    I know what you want to say.
    For part ii, where you have an obtuse angle, I would use the fact that sinb=3/7 and that sin^2+cos^2=1
    I could find cos^2 and then cos. I would get 2 values. I would chose the negative one since cos is negative in the 2nd quarter.
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    (Original post by purplemind)
    If angle is obtuse, it is more than 90 and less than 180.
    From CAST diagram you know, that in the 2nd quarter (more than 90, less than 180, so that's an obtuse angle) sin is positive, cos is negative and tan is negative. This should help.
    thanks it does help
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    (Original post by BenChard)
    I still have no idea here as the angles in a triangle must add up to 180 degrees. right angle=90, obtuse=90+ therefore the other angle is non-existent so we cant use pythag?
    You might want to think of it like this. Let  \theta be the acute angle with the same sine as beta. Then:

     sin \theta = \frac{3}{7}

     cos \theta = \frac{\sqrt{40}}{7}

    By symmetry of the sine graph,  \beta = 180 ^\circ - \theta

    So we are looking for:

     cos (180^\circ - \theta) = cos(-(\theta - 180 ^\circ)) = cos(\theta - 180 ^\circ) = -cos(\theta) = -\frac{\sqrt{40}}{7}

    This is using pythagoras and a bit of visualizing graphs until you get used to these rules.
 
 
 
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