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    Attachment 728206
    I can't get the right answer
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    (Original post by Sammyboy1234)
    Attachment 728206
    I can't get the right answer
    So then post your working out and we'll see where you're going wrong.

    EDIT: The angle between the two is not 55. It's 90+35
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    all I have done is used the cosine rule and gotten 281 for the distance
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    (Original post by Sammyboy1234)
    all I have done is used the cosine rule and gotten 281 for the distance
    Yes this is clear, but what numbers did you use for the cosine rule.

    Also, the attachment is gone.
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    35 squared +270 squared - 2 ×35×270×cos105
    then squared root 79016....
    gives you 281
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    (Original post by Sammyboy1234)
    ...
    Firstly, when using a^2+b^2-2ab \cos \theta, note that a,b are actually the lengths of the lines. Right now you have these lengths labelled as their bearings for some reason. Secondly, \theta is the angle between the two arrows - it is not 105.

    Lastly, we are interested in the distance between the two objects AFTER 2 hours. So when it comes to a,b values, they need to represent distances of both ships from the Port after 2 hours.
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    would I need to use the speed distance over time formula
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    (Original post by Sammyboy1234)
    would I need to use the speed distance over time formula
    To work out how far away each ship is from the Port, yes.

    \text{speed} \times \text{time} = \text{distance}
 
 
 
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