Further Maths - Minimum point of a quadratic

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    I can't figure out how to work out question 5 (a).

    So far, I have the basics;
    PCA = x
    CPB = 2x

    I don't know what to do after this...

    http://www.mrbartonmaths.com/resourc.../Paper%202.pdf
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    (Original post by RosaA)
    I can't figure out how to work out question 5 (a).

    So far, I have the basics;
    PCA = x
    CPB = 2x

    I don't know what to do after this...

    http://www.mrbartonmaths.com/resourc.../Paper%202.pdf
    No, \mathbf{CPB}=180-2x as it's an isosceles triangle.

    Angles on a straight line add up to 180 therefore \mathbf{BPC}+(180-2x)=180 and work out BPC. Then use it's an isosceles triangle and use that to your advantage for ABC.
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    (Original post by RDKGames)
    No, \mathbf{CPB}=180-2x

    Angles on a straight line add up to 180 therefore \mathbf{BPC}+(180-2x)=180 and work out BPC. Then use it's an isosceles triangle and use that to your advantage for ABC.
    Ah, okay thanks. I thought I has some idea about what I was doing by patently not!

    Thank you for the help
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    (Original post by RosaA)
    Ah, okay thanks. I thought I has some idea about what I was doing by patently not!

    Thank you for the help
    No problem, though this question has nothing to do with minimum points let alone quadratics as the title suggests.
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    (Original post by RDKGames)
    No problem, though this question has nothing to do with minimum points let alone quadratics as the title suggests.
    I know, I realised after posting that I had written an incorrect title.

    In terms of the question, I'm still a bit confused to be perfectly honest.

    180 = 2x - 2x + 180
    APC = 180 - 2x ??

    Do you mind just going through it because I am completely losing the plot
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    (Original post by RosaA)
    I know, I realised after posting that I had written an incorrect title.

    In terms of the question, I'm still a bit confused to be perfectly honest.

    180 = 2x - 2x + 180
    APC = 180 - 2x ??

    Do you mind just going through it because I am completely losing the plot
    Triangles APC and BPC are both isosceles due to equal sides. This is useful.

    For triangle APC, we know that angle \mathbf{PAC}=x therefore \mathbf{PCA}=x as well. Since angles in a triangle add up to 180 this means that 180=\mathbf{PAC}+ \mathbf{PCA} + \mathbf{CPA} = x + x + \mathbf{CPA} \Rightarrow \mathbf{CPA}=180-2x

    Angles on a straight line add up to 180, therefore on the line BA: 180=\mathbf{CPA}+\mathbf{BPC}=(1  80-2x)+\mathbf{BPC}

    You can find angle \mathbf{BPC} in terms of x therefore find angle \mathbf{ABC} as it is an isosceles triangle.
 
 
 
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