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    (Original post by I hate maths)
    I'll challenge that claim with my own improved method.

    Draw the diagram to scale and use a protractor.
    Wouldn't that be impossible since pi is irrational?
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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
    I'd recommend following the bear's method because it's the most GCSE and ask him if you get stuck.
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    (Original post by the bear)
    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.
    But if we already
    (Original post by the bear)
    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.
    Is this right?
    I don’t know what to do
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    (Original post by Y11_Maths)
    But if we already

    Is this right?
    I don’t know what to do
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    Your last line is wrong. You have r^2 on both sides so they should cancel.
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    (Original post by Notnek)
    Your last line is wrong. You have r^2 on both sides so they should cancel.
    Oh yes, silly mistake, this time of night haha.
    So 1=2sinx.
    Where do I go from here? I don’t see where it’s taking me...
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    (Original post by Y11_Maths)
    Oh yes, silly mistake, this time of night haha.
    So 1=2sinx.
    Where do I go from here? I don’t see where it’s taking me...
    You have 1=2sin(2x) - I assume you meant this?

    Move the 2 to the other side then you should know what 2x is equal to.
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    (Original post by SumOfSquares)
    They did cancel. I think it says 1 on the LHS
    No I shouldn’t be left with an r^2 term on the RHS
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    (Original post by Y11_Maths)
    No I shouldn’t be left with an r^2 term on the RHS
    Yh. I just realized after posting. Sorry.
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    (Original post by SumOfSquares)
    Wouldn't that be impossible since pi is irrational?
    I was messing around . Standards for proof in school maths are lax but not THAT lax. I think there are other issues if you were to approach the problem that way... You'd have to take a long good look at yourself in the mirror!
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    (Original post by Notnek)
    You have 1=2sin(2x) - I assume you meant this?

    Move the 2 to the other side then you should know what 2x is equal to.
    Yes I did my bad. So sin(2x)= 1/2. Which is 30 degrees. So the other angle is 150 degrees. Since the triangle is isosceles 2x is 30 so x is 15 degrees! Please tell me that I’ve finished????
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    (Original post by Y11_Maths)
    Yes I did my bad. So sin(2x)= 1/2. Which is 30 degrees. So the other angle is 150 degrees. Since the triangle is isosceles 2x is 30 so x is 15 degrees! Please tell me that I’ve finished????
    Looks like you've finished. Although once you know 2x = 30, you have x = 15 and you're done - I don't know why you mentioned isosceles triangles.

    For GCSE you need to know that there are two solutions to sin(x) = a where x<180 and the second can be found by subtracting the first from 180. So another solution is 2x = 150 so x = 75. But your teacher probably doesn't expect you to mention this.
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    (Original post by Notnek)
    Looks like you've finished. Although once you know 2x = 30, you have x = 15 and you're done - I don't know why you mentioned isosceles triangles.

    For GCSE you need to know that there are two acute solutions to sin(x) = a and the second can be found by subtracting the first from 180. So another solution is 2x = 150 so x = 75. But your teacher probably doesn't expect you to mention this.
    Well if triangle OCA has 2 lengths of r then it is isosceles and has 2 angles both of x. Which is why I said this. But I’m tired so I missed your quicker way but it’s fine.

    Oh ok thanks for bringing this up! And thank you for helping. Also huge thanks to psc---maths for providing the diagram, this was extremely useful! And thanks to the bear who made it really simple to follow and get the answer. Wohoo!
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    My solution:

    From the given statement, area of triangle = r^2/2.

    Connecting radii helps to see area of triangle = 1/2 r^2(sina) + 1/2 r^2 sin(180-x) --> [ABC] = r^2sin(a). This means a=30 degrees. <X = 0.5(180-150) = 15 degrees (as the triangle is evidently isosceles).
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    (Original post by thekidwhogames)
    My solution:

    From the given statement, area of triangle = r^2/2.

    Connecting radii helps to see area of triangle = 1/2 r^2(sina) + 1/2 r^2 sin(180-x) --> [ABC] = r^2sin(a). This means a=30 degrees. <X = 0.5(180-150) = 15 degrees (as the triangle is evidently isosceles).
    Edit: had to use sin(x) = sin(180-x) but this is in the new GCSE :P
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    (Original post by Y11_Maths)
    And thanks to the bear who made it really simple to follow and get the answer. Wohoo!
    Woohoo

    :naughty:
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    (Original post by the bear)
    Woohoo

    :naughty:
    Are there any limits to the bear’s helpfulness?
 
 
 
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