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    Abcd is a rectangle and I need to find out the value of x when the length is 18 cm and the width is 14cm (bottom)
    It has two diagonal lines inside te rectangle dividing it into 4 triangles one labelled x
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    Theres a photo attached
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    (Original post by Forevereuphoria)
    Theres a photo attached
    No there's not.
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    (Original post by mackemforever)
    No there's not.
    There is now
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    Image is attached
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    (Original post by mackemforever)
    No there's not.
    Yes there is.

    I'll take a look.
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    (Original post by Forevereuphoria)
    Theres a photo attached
    What have you tried so far? Can you take a picture of your working and we can see where you need to go next?
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    (Original post by Forevereuphoria)
    Theres a photo attached
    X= angle CXD
    tanC = 18/14
    Length BD=length AC (use Pythagoras)
    Divided by two, that length is equal to DX
    you can then use the sin rule sinX/14 = sinC/Length DX to find x

    actually, that seems a lot for 3 marks, there's probably a better way
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    (Original post by shutupem)
    X= angle CXD
    tanC = 18/14
    Length BD=length AC (use Pythagoras)
    Divided by two, that length is equal to DX
    you can then use the sin rule sinX/14 = sinC/Length DX to find x

    actually, that seems a lot for 3 marks, there's probably a better way
    I still dont understand could you explain what you are talking about i.e why its tan? Put examples in
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    Image is attached for both please explain
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    (Original post by Forevereuphoria)
    I still dont understand could you explain what you are talking about i.e why its tan? Put examples in
    What have you done so far?
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    it's just 90 degree trig...SOHCAHTOA

    PS your picture is upside down
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    (Original post by Slowbro93)
    What have you done so far?
    Sorry didn't notice your question I've done nothing that looks correct so theres no point me posting it.
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    (Original post by Forevereuphoria)
    I still dont understand could you explain what you are talking about i.e why its tan? Put examples in
    Opposite angles are equal, so x is angle CXD
    ACD is a right angled triangle
    You have 14 and 18, two sides which are not the hypotenuse.

    You can work out the size of angle C, which is the same in the right angled triangle and in the triangle containing angle CXD, using SohCahToa.

    The two sides you know he lengths of are the opposite and adjacent sides, relative to angle C, so you use tan(angle) = opp/adj.

    If you knew the length of the hypotenuse, so you could do Pythagoras first instead of second, then you could use sin or cos. I did the SohCahToa first, so had to use tan as those were the only two sides I knew the length of.

    Hope that helps.
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    (Original post by shutupem)
    Opposite angles are equal, so x is angle CXD
    ACD is a right angled triangle
    You have 14 and 18, two sides which are not the hypotenuse.

    You can work out the size of angle C, which is the same in the right angled triangle and in the triangle containing angle CXD, using SohCahToa.

    The two sides you know he lengths of are the opposite and adjacent sides, relative to angle C, so you use tan(angle) = opp/adj.

    If you knew the length of the hypotenuse, so you could do Pythagoras first instead of second, then you could use sin or cos. I did the SohCahToa first, so had to use tan as those were the only two sides I knew the length of.

    Hope that helps.
    im still baffled...
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    (Original post by Forevereuphoria)
    im still baffled...
    What about it is confusing?
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    (Original post by shutupem)
    What about it is confusing?
    How are you labelling the sides...
    is the answer 90
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