h26
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I have got the first part so k=6 but the rest is rather confusing
Can someone help please?
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username1732133
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(Original post by h26)
I have got the first part so k=6 but the rest is rather confusing
Can someone help please?
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Consider P as the point (6 \cos \theta, 6 \sin \theta).
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h26
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(Original post by BuryMathsTutor)
Consider P as the point (6 \cos \theta, 6 \sin \theta).
Thanks very much for the reply - but I was wondering if you could please kindly explain it in simple terminology in the way the mark scheme puts it sorry it probably is in simple terms but I still don't quite understand that bit. I have attached the mark scheme below:

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So I have got up to k=6 and then the rest really baffles me
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(Original post by h26)
Thanks very much for the reply - but I was wondering if you could please kindly explain it in simple terminology in the way the mark scheme puts it sorry it probably is in simple terms but I still don't quite understand that bit. I have attached the mark scheme below:

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So I have got up to k=6 and then the rest really baffles me
In the diagram that you attached there is a right-angled triangle.

The angle at the centre is marked alpha. For this angle, what are the lengths of the opposite side and the adjacent side?

Can you use these two lengths to work out alpha?
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(Original post by BuryMathsTutor)
In the diagram that you attached there is a right-angled triangle.

The angle at the centre is marked alpha. For this angle, what are the lengths of the opposite side and the adjacent side?

Can you use these two lengths to work out alpha?
Thanks very much, so 6costheta = 3 --> theta =60 degrees
6sintheta = (27)^1/2 --> theta = 60 degrees
However, wouldn't this be classified as C4 Maths? This is a C1 question from the MEI website, so this was confusing as I never apply C4 Maths in a C1 question.

Also, what is the purpose of the right-angled triangle in the mark scheme? From the question, I know triangle PQR is equilateral, so all three of its angles are 60degrees. Therefore, I find the use of the right-angled triangle confusing as triangle PQR is equilateral.
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(Original post by h26)
Thanks very much, so 6costheta = 3 --> theta =60 degrees
6sintheta = (27)^1/2 --> theta = 60 degrees
However, wouldn't this be classified as C4 Maths? This is a C1 question from the MEI website, so this was confusing as I never apply C4 Maths in a C1 question.

Also, what is the purpose of the right-angled triangle in the mark scheme? From the question, I know triangle PQR is equilateral, so all three of its angles are 60degrees. Therefore, I find the use of the right-angled triangle confusing as triangle PQR is equilateral.
The bit of maths you're using there is often referred to SOHCAHTOA, at GCSE. The right-angled triangle is used to work out the angle alpha. Once you have that you can work out the coordinates of the other vertices.
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(Original post by BuryMathsTutor)
The bit of maths you're using there is often referred to SOHCAHTOA, at GCSE. The right-angled triangle is used to work out the angle alpha. Once you have that you can work out the coordinates of the other vertices.
Thanks very much for your help so far, it has cleared up a lot of things. But how do you use the angle alpha to work out the coordinates of the other vertices?

Also, in the markscheme, it says "The three points are spaced out equally around the circle, so the angles at the centre are 120 degrees" and is then straight away followed by "So for an equilateral triangle the vertices Q and R are (3, -√27) and (-6,0)." I am not sure what they are implying by the angles at the centre and how it totals to 120 degrees. I am also not sure how this gives the vertices of the triangle.

Sorry for my ignorance.
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(Original post by h26)
Thanks very much for your help so far, it has cleared up a lot of things. But how do you use the angle alpha to work out the coordinates of the other vertices?

Also, in the markscheme, it says "The three points are spaced out equally around the circle, so the angles at the centre are 120 degrees" and is then straight away followed by "So for an equilateral triangle the vertices Q and R are (3, -√27) and (-6,0)." I am not sure what they are implying by the angles at the centre and how it totals to 120 degrees. I am also not sure how this gives the vertices of the triangle.

Sorry for my ignorance.
There's no need to apologise.
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