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    (Original post by Y11_Maths)
    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
    Excuse my lack of ruler/ ability in drawing diagrams properly. I tend not to put in too much effort into diagrams unless I am doing an actual exam!

    Have you encountered the fact that the angle at the centre of the circle is twice the angle at the circumference? From there, we are using the fact that AB is a straight line and so angle AOC+ angle COB = 180.
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    (Original post by Y11_Maths)
    Isn’t the area of the triangle r^2/2, or am I wrong?
    No, the area of the triangle has x in there somewhere, else how are you going to figure out what x is?
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    (Original post by I hate maths)
    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.
    I asked if you could substitute x as 15 degrees in and he said no, you have to prove that it is the final outcome not that everything else fits perfectly into place when that value is 15 degrees...
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    (Original post by psc---maths)
    Excuse my lack of ruler/ ability in drawing diagrams properly. I tend not to put in too much effort into diagrams unless I am doing an actual exam!

    Have you encountered the fact that the angle at the centre of the circle is twice the angle at the circumference? From there, we are using the fact that AB is a straight line and so angle AOC+ angle COB = 180.
    Yes I understand this but don’t know where you got your values from
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    (Original post by I hate maths)
    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.
    This assumes it's a calculator question. The method given by psc---maths is a bit nicer for GCSE since you can use sin(30) = 1/2. Although it relies on sin(2x) = sin(180-2x) which I wouldn't expect most GCSE students to realise.

    If it was a calculator question then since it's a "show that", you could just set x as 15 and find the areas in terms of r.
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    (Original post by Y11_Maths)
    I asked if you could substitute x as 15 degrees in and he said no, you have to prove that it is the final outcome not that everything else fits perfectly into place when that value is 15 degrees...
    I don't mean subbing x=15 straight away into the diagram and working with that throughout. It's fine to work with x's all the way through and get to the last stage where "expression in x"="something" and show that x=15 satisfies that condition, and that it's the only value which could satisfy that condition (in this case, it's not though). It's something to at least be aware of. Psc's diagram will provide a nice direct method, and probably the way your teacher wants you to approach the problem, so I'd go with that certainly.

    EDIT: Notnek you're completely right, I assumed this is a calculator question. Forgot non-calculator GCSE papers are a thing... My bad.
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    I’m woeful at maths, and I have no idea what any of this means or what any of you are saying. Nonetheless I’ve read the whole thing!
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    angle COB = 2x

    the area of triangle ABC = r2/2

    the area of triangle OBC = 1/2 of the area of triangle ABC

    ====> r2/4 = 1/2*OB*OC*Sin2x

    etc.
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    (Original post by psc---maths)
    Excuse my lack of ruler/ ability in drawing diagrams properly. I tend not to put in too much effort into diagrams unless I am doing an actual exam!

    Have you encountered the fact that the angle at the centre of the circle is twice the angle at the circumference? From there, we are using the fact that AB is a straight line and so angle AOC+ angle COB = 180.
    Is this right? Sorry if the image quality is poor
    Name:  E6ACB29C-A0DB-49F7-B252-2082FFE8224A.jpg.jpeg
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Size:  24.5 KB
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    (Original post by Notnek)
    Pour quoi?
    so we are told that the triangle area T / the circle area = 1/(2π)

    so T = πr2/(2π)

    T = r2/2
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    (Original post by the bear)
    so we are told that the triangle area T / the circle area = 1/(2π)

    so T = πr2/(2π)

    T = r2/2
    That’s how I did it too
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    (Original post by the bear)
    so we are told that the triangle area T / the circle area = 1/(2π)

    so T = πr2/(2π)

    T = r2/2
    Why is the area of triangle OBC = 1/2 of the area of triangle ABC?
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    (Original post by Y11_Maths)
    That’s how I did it too
    :five:

    http://weknowmemes.com/generator/upl...7455664152.jpg
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    (Original post by Notnek)
    Why is the area of triangle OBC = 1/2 of the area of triangle ABC?
    triangles OBC and OBA both have the same base & perpendicular height.
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    I haven't tried it, but I think you can solve using simultaneous equations.
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    Aaah I’m so confused. There are so many ways of going about this problem and I don’t even know which is right
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    (Original post by Y11_Maths)
    Aaah I’m so confused. There are so many ways of going about this problem and I don’t even know which is right
    Which part is confusing?
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    (Original post by the bear)
    triangles OBC and OBA both have the same base & perpendicular height.
    Indeed they do. Your method is the most GCSE of the thread
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    (Original post by Notnek)
    Indeed they do. Your method is the most GCSE of the thread
    I'll challenge that claim with my own improved method.

    Draw the diagram to scale and use a protractor.

    Spoiler:
    Show
    ha take that bear
 
 
 
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