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Math Help - Easy Trigonometry Problem

  1. #1
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    Easy Trigonometry Problem

    A yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by 2 km due west.

    a) How long is the final leg?


    I drew the diagram and got 3.2 km. However the answer is supposed to be 5.39 km, though it could be wrong.

    Any help will be appreciated!
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    A yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by 2 km due west.

    a) How long is the final leg?


    I drew the diagram and got 3.2 km. However the answer is supposed to be 5.39 km, though it could be wrong.

    Any help will be appreciated!
    I would use vectors to solve this problem rather than trig (apart from finding direction)

    Let i be 1km due east and let j be 1km due north. Also let the origin be the start point

    If we say his route is r then, relative to the origin his position will be the sum of the vectors :

    r = 4i + (-5j) + (-2i) = 2i - 5j

    Spoiler:
    |r| = \sqrt{2^2 + (-5)^2} = \sqrt{29} = 5.39km

    (note that r in the final step should be in bold but I don't know that particular LaTeX code )

    The direction is given by tan(\frac{-5}{2}) and this angle is to the horizontal in the 4th quadrant. From due north this will be

    \frac{\pi}{2} + tan(\frac{-5}{2}) or

    180 + tan(\frac{-5}{2}) if you prefer degrees
    Last edited by e^(i*pi); April 21st 2009 at 09:27 AM. Reason: Added in direction part (and spoiler)
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    Exclamation

    I'm sorry to be so picky...but do you think you can solve it using trigonometry only. I haven't gone through vectors, I think it would be more beneficial for me if you use trigonometry only. I don't understand the above.

    Thank you.
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    Quote Originally Posted by Joker37 View Post
    A yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by 2 km due west.

    a) How long is the final leg?


    I drew the diagram and got 3.2 km. However the answer is supposed to be 5.39 km, though it could be wrong.

    Any help will be appreciated!
    See attached.
    Attached Thumbnails Attached Thumbnails Easy Trigonometry Problem-a1.jpg  
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  5. #5
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    Talking

    Quote Originally Posted by masters View Post
    See attached.
    Of course! Should have seen that. Thank you very much!
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    Quote Originally Posted by Joker37 View Post
    I'm sorry to be so picky...but do you think you can solve it using trigonometry only. I haven't gone through vectors, I think it would be more beneficial for me if you use trigonometry only. I don't understand the above.

    Thank you.
    Sure.

    If you draw a line from the end of the third stage vertically up to the line representing the first stage and draw in the hypotenuse there should be a right-angled triangle. (See attachment)

    To find the total distance of the last leg use Pythagoras' Theorem to get the hypotenuse which should be \sqrt{29} which is roughly 5.39km

    You can then use tan(\theta) = \frac{-5}{2}

    \theta = arctan(\frac{-5}{2})

    As all angles in a triangle add up to 180^o the final angle is:

    \alpha = 180 - (90+ arctan(\frac{-5}{2}))
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    Okay last question! Sorry usually I would start a new thread if I were to introduce another question however, this one's an easy one and it is linked to the top post.

    b) On what bearing must the final leg be sailed?


    Thanks.
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  8. #8
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    Quote Originally Posted by Joker37 View Post
    Okay last question! Sorry usually I would start a new thread if I were to introduce another question however, this one's an easy one and it is linked to the top post.

    b) On what bearing must the final leg be sailed?


    Thanks.
    Quote Originally Posted by e^(i*pi) View Post
    tan(\theta) = \frac{5}{2}
    From mine or masters' diagram (which is cleaner than mine xD) the angle between the horizontal and the end of the third leg (theta on my diagram) is given above using the quote from my last post. However, as all bearings are from due north you will have to add 270 degrees from due north to the horizontal.
    Last edited by e^(i*pi); April 21st 2009 at 10:12 AM.
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  9. #9
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    Quote Originally Posted by Joker37 View Post
    Okay last question! Sorry usually I would start a new thread if I were to introduce another question however, this one's an easy one and it is linked to the top post.

    b) On what bearing must the final leg be sailed?


    Thanks.
    See diagram
    Attached Thumbnails Attached Thumbnails Easy Trigonometry Problem-a1.jpg  
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