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Fp1 exam question help

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    question 7 jan 09 fp1 aqa
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    (link if wanted) http://store.aqa.org.uk/qual/gceasa/...W-QP-JAN09.PDF

    i understand the theory i.e i have to prove it is linear interpolation however i can't seem to obtain the rearranged version for r - mainly the c-d part why is that needed and not just d for the similar triangles?, could someone walk it through please.

    Also, on q3 am i right in thinking i have too minus the (pi divided by 2) then divide by -3?
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    Oh yeah ... horrible question
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    (Original post by TenOfThem)
    Oh yeah ... horrible question
    so i shouldn't feel that annoyed that i was stuck and spent a lot of time on this?
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    I seem to remember that I spent ages doing convoluted adding of stuff to make it work

    Then when I was showing the class I saw a really obvious way
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    (Original post by coolstorybrother)
    so i shouldn't feel that annoyed that i was stuck and spent a lot of time on this?
    I spent time using the little triangles .... then in class I thought ... hang on and used this

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    (Original post by TenOfThem)
    I seem to remember that I spent ages doing convoluted adding of stuff to make it work

    Then when I was showing the case I saw a really obvious way
    im not sure but is the really obvious way using similar triangles (r-a)/c = (b-r)/d ? could you help on q3?
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    (Original post by coolstorybrother)
    im not sure but is the really obvious way using similar triangles (r-a)/c = (b-r)/d ? could you help on q3?
    It is about using the correct similar triangles

    see ^^^^^^
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    What is arctan(rt3)
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    (Original post by TenOfThem)
    I seem to remember that I spent ages doing convoluted adding of stuff to make it work

    Then when I was showing the case I saw a really obvious way
    Which part of the question is difficult?
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    (Original post by raheem94)
    Which part of the question is difficult?
    none of it
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    (Original post by TenOfThem)
    none of it
    On which part did you spent ages?
    I may like to have a go at it.
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    (Original post by raheem94)
    On which part did you spent ages?
    I may like to have a go at it.
    The initial proof

    As I said I used the 2 small triangles and then had to do manipulation to get the format required

    But ... as my diagram shows it was very easy
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    (Original post by TenOfThem)
    The initial proof

    As I said I used the 2 small triangles and then had to do manipulation to get the format required

    But ... as my diagram shows it was very easy
    I did the 1st part in this way,

    The gradient of the line(using points P(a,c) and Q(b,d) ) is,  \displaystyle m = \frac{d-c}{b-a}

    Now form the equation of the line, using the point R(r,0),
     \displaystyle y - 0 = \frac{d-c}{b-a} (x-r) \implies y(b-a)  = (d-c)(x-r) \\ \implies y(b-a) = dx - dr -cx +cr  \implies y(b-a) = x(d-c) + r(c-d) \\ \implies y(b-a) = -x(c-d) + r(c-d) \implies y \left( \frac{b-a}{c-d} \right) = -x + r \\ \implies r = x + y \left( \frac{b-a}{c-d} \right)

    Now sub in P(a,c), this gives,

     \boxed{ r = a + c \left( \frac{b-a}{c-d} \right)  }

    How is this solution?
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    I did Purple ratio = Green ratio

    \dfrac{r-a}{c} = \dfrac{b-a}{c-d}

    so

    r-a = c(\frac{b-a}{c-d})

    so

    r = a + c(\frac{b-a}{c-d})
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    (Original post by raheem94)

    How is this solution?

    I did similar to this before I realised the easy way I have shown above

    That is what I meant when I said it took too much manipulation
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    (Original post by TenOfThem)
    I did similar to this before I realised the easy way I have shown above

    That is what I meant when I said it took too much manipulation
    Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

    Though my way is also not very long, it only takes around a minute.
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    (Original post by raheem94)
    Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

    Though my way is also not very long, it only takes around a minute.
    Yeah ... from experience I just knew it should be "see-able" rather than "workable"


    Once I "saw" it I was like "OMG how obvious"
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    (Original post by TenOfThem)
    It is about using the correct similar triangles

    see ^^^^^^
    i kind of understand this... everything except why it is c-d?

    (Original post by TenOfThem)
    What is arctan(rt3)
    umm i really dont know, should i?
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    (Original post by coolstorybrother)
    i kind of understand this... everything except why it is c-d?



    umm i really dont know, should i?
    c-d ... yeah I had to think ... but d is already a negative number


    Yes you should
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    (Original post by raheem94)
    Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

    Though my way is also not very long, it only takes around a minute.
    Similar triangles is widely used as a method for linear-interpolation.

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