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    So my maths teacher gave me this question at the end of school today and I told him that I would get it done for tomorrow. He is currently "one over on me" so I really want to get the answer back to him so we are even and I have the last laugh. However, the question he gave me has seriously perplexed me and I do not know what to do.

    I worked out that bh=r^2 so the area of the triangle is r^2/2 and the area of the circle is pi r^2. But this has taken me nowhere...

    Any suggestions?
    Thanks in advance

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    Hmm the challenge is to do this without using A Level methods...
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    (Original post by Notnek)
    Hmm the challenge is to do this without using A Level methods...
    So is this a GCSE question?
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    (Original post by Y11_Maths)
    So is this a GCSE question?
    This is definitely doable for GCSE students, but it requires some thought.
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    (Original post by Radioactivedecay)
    This is definitely doable for GCSE students, but it requires some thought.
    Help, please?
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    I'm attempting it...
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    (Original post by Y11_Maths)
    Help, please?
    Try working out CA by splitting the triangle and somehow forming a right angled triangle, then use 1/2ABsinC for the area, then you should see it from there.
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    Name:  871AC4A5-6014-4912-8465-6FA13C6A4BB6.jpg.jpeg
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    Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!
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    (Original post by Notnek)
    Hmm the challenge is to do this without using A Level methods...
    Which bit in your work did you need A Level maths? If you did something similar to me, you don't actually need to simplify the trig with A Level identities but you could just sub the answer given in to confirm it's the solution.

    My gripe with the question is that it doesn't specify angle BAC is less than CBA, so BAC could be 75 instead.
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    (Original post by psc---maths)
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    Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!
    I don't think that diagram is correct.
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    (Original post by I hate maths)
    Which bit in your work did you need A Level maths? If you did something similar to me, you don't actually need to simplify the trig with A Level identities but you could just sub the answer given in to confirm it's the solution.
    Yes you can just sub it in but I was looking for a nicer GCSE solution that gets the answer directly. You wouldn't see a question like this in a GCSE paper.
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    (Original post by psc---maths)
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    Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!
    The height isn’t r though?
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    (Original post by Y11_Maths)
    The height isn’t r though?
    It isn't meant to be the height. It is the line drawn from the centre to B, so it is a radius of length r
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    (Original post by psc---maths)
    It isn't meant to be the height. It is the line drawn from the centre to B, so it is a radius of length r
    I'm suggesting you use 1/2absinc for both of the smaller triangles
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    (Original post by Y11_Maths)
    The height isn’t r though?
    Disregard my earlier comment. An easier way would be find the other two sides of the triangle in terms of r and x using SOCAHTOA, and then finding the area of the triangle from there.
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    (Original post by Radioactivedecay)
    I don't think that diagram is correct.
    I think it is correct but the original triangle is easy enough to work with without splitting.
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    (Original post by I hate maths)
    I think it is correct but the original triangle is easy enough to work with without splitting.
    Wrt the other way would that not require 2sinxcosx=sin2x. I was giving a gcse level answer
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    (Original post by psc---maths)
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    Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!
    The diagram doesn’t look accurate? Where did you get 180-2x from? The top vertice of the triangle is not vertically above the centre and how can OC be r? I’m really confused
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    (Original post by Radioactivedecay)
    Disregard my earlier comment. An easier way would be find the other two sides of the triangle in terms of r and x using SOCAHTOA, and then finding the area of the triangle from there.
    Isn’t the area of the triangle r^2/2, or am I wrong?
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    (Original post by psc---maths)
    Wrt the other way would that not require 2sinxcosx=sin2x. I was giving a gcse level answer
    As I was saying to Notnek, it wouldn't require this as the question gives you the value of x, so you can sub it in and confirm it's the solution without knowledge of identities. People tend to have an aversion for this kind of solution but I think it's a useful way to do some problems.
 
 
 
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