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Further Maths - Minimum point of a quadratic

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/resources/GCSE%20Revision/Further%20Maths%20Past%20Papers/5.%20June%202012/Paper%202.pdf
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/resources/GCSE%20Revision/Further%20Maths%20Past%20Papers/5.%20June%202012/Paper%202.pdf


No, CPB=180−2x\mathbf{CPB}=180-2x as it's an isosceles triangle.

Angles on a straight line add up to 180 therefore BPC+(180−2x)=180\mathbf{BPC}+(180-2x)=180 and work out BPC. Then use it's an isosceles triangle and use that to your advantage for ABC.
(edited 7 years ago)
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RosaA
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Original post by RDKGames
No, CPB=180−2x\mathbf{CPB}=180-2x

Angles on a straight line add up to 180 therefore BPC+(180−2x)=180\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
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. :smile:
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RosaA
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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. :smile:


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 :s-smilie:
(edited 7 years ago)
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 :s-smilie:


Triangles APC and BPC are both isosceles due to equal sides. This is useful.

For triangle APC, we know that angle PAC=x\mathbf{PAC}=x therefore PCA=x\mathbf{PCA}=x as well. Since angles in a triangle add up to 180 this means that 180=PAC+PCA+CPA=x+x+CPA⇒CPA=180−2x180=\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=CPA+BPC=(180−2x)+BPC180=\mathbf{CPA}+\mathbf{BPC}=(180-2x)+\mathbf{BPC}

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

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