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Math Help - Is there enough info given in question

  1. #1
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    Question Is there enough info given in question

    A ship which leaves a port, p, and travels south west, with constant speed, observes the flash of a lighthouse, l, ina direction S60W. After travelling for 10km, to a point,t, the ship observes the flash of the same lighthouse, due west.

    Q1)Calculate how far the ship is from the lighthouse at this time i.e calculate length tl.

    The answer given for this is 5.18km- Having sketched the diagram, I can't see that enough details have been supplied to use sine rule or otherwise to calculate this.

    Q2) Calculate how close to the lighthouse the ship will pass.

    The answer given here is 3.66km. Once again I cant figure this out. I took the point at which the ship passes the lighthouse to be the point at which t moves under the point l and sketched the resultant triangle. I used the answer from the first part to solve the triangle but had no joy.

    Any ideas??
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  2. #2
    MHF Contributor Unknown008's Avatar
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    If you make your sketch, and label the points 1: the point the ship is when it first see the flash, 2 where it sees the flash for the second time and L the position of the lighthouse, you get:

    Angle L12 = 15 degrees
    Angle 1L2 = 90 - 60 = 30 degrees
    Hence, angle 12L = 180 - (60 + 15) = 105 degrees

    Side 12 = 10 km

    Use the sine rule.

    Let D be the distance you're looking for, that is when the ship is at 1.

    \dfrac{D}{\sin(105)} = \dfrac{10}{\sin(30)}

    Get this one first.
    Last edited by Unknown008; February 2nd 2011 at 10:26 AM. Reason: Corrected angle 1L2, see post below. Also, I think that my D here is something else, oops...
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Angle 1L2 = 90 - 30 = 60 degrees
    How so? I'm getting this, with your labels...



    ... and therefore \dfrac{D}{\sin(15)} = \dfrac{10}{\sin(30)}

    which is coming out right.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oops, brain fart, I thought 60 degrees W of south means the upper angle 30 degrees, then 60 degrees will be the lower angle without... well, you see
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  5. #5
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    Post Where are you getting the 15 degree angle from? Thanks, by the way!

    Quote Originally Posted by tom@ballooncalculus View Post
    How so? I'm getting this, with your labels...



    ... and therefore \dfrac{D}{\sin(15)} = \dfrac{10}{\sin(30)}

    which is coming out right.

    How do you get the 15 degree angle? Can work it out from there. Thanks for the help.
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  6. #6
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    Sw = S45W
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  7. #7
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    Oh, should have realised that! Was driving me mad! Thanks again.
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