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Math Help - Is this method effective enough or is there better?.

  1. #1
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    Is this method effective enough or is there better?.

    Hi,

    i'm measuring the distance of a boat moored on a 22.5 degree angle being the front nose hard up against the mooring and rear end hanging out on the 22.5 degree angle, i want to know the distance from the mooring of the rear end on the 52 foot boat.

    i come up with this model/formula is it right, is there other ways, i know very little about maths.

    i worked out (i think) using a model, i took a room 3.5 m X 3.5 m went from one corner diagonally to another corner that gave me the 45 degrees cutting that in half gave me 22.5 degrees? measuring half distance from wall to wall, and being the distance from the 0-180 degree wall line at the 22.5 degree mark was for this scenario 1.75 metres.
    being 3500mm (3.5m) and the boat being 15849.6mm (15.84+m) and dividing that with each other gave me 4.52845714+ and times by 1750mm (1.75m)(22.5 degrees) gives me 7.9248m or 79248mm, being the distance (excluding the width of the boat which would add that distance on top of the 7.9m being the further-est distance) of the rear end of the boat away from the mooring, is this right?. or am i wrong, is there an easier way i'm not the worlds best at maths, amerature!.

    any help welcome.
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  2. #2
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    Re: Is this method effective enough or is there better?.

    If you send a straight line from the end of the boat to the mooring you've completed a right triangle. One of the trig relationships we have is that the sine of an angle is equal to the ratio of the length of the side opposite the angle to the hypotenuse of the triangle. That is, \sin \theta = \frac{opposite}{hypotenuse}. For your problem the opposite length will be the distance from the end of the boat to the mooring while the length of the boat is the hypotenuse. This works out to 52 * \sin 22.5 \approx 20 feet
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  3. #3
    Odo
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    Re: Is this method effective enough or is there better?.

    I agree with the answer given by cshanholtze (which was posted at the same time I was doing the drawing in Word...), that is the correct way of doing it.

    I made a figure to illustrate, and make sure that you are interested in finding X, or the length of the segment BC.

    Is this method effective enough or is there better?.-boat_problem.png

    In my figure I have the angle denoted by ALPHA, and cshanholtze has given you the sine formula using THETA, which is saying the exact same thing!

    X=52 * sin(22.5deg) = 19.8995...ft
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  4. #4
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    Re: Is this method effective enough or is there better?.

    thanx heaps for the replys, much appreciated.

    if you can and time permits and as time not being at all on my side as i have very little time to look around on the site/web, can you direct me in the direction of how to go about learning maths, i know very little.

    thanx
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  5. #5
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    Re: Is this method effective enough or is there better?.

    i think what you have mentioned above might work
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