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Math Help - Law of Sine the Ambiguous Case Problems

  1. #1
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    Law of Sine the Ambiguous Case Problems

    I am super confused about these-
    I have two problems that I have no idea what to do with. I know I need to draw a picture but I do not understand the problems enough to draw the diagram. For each I will type out the problem as well as attach an image of the problems as I am doing them on the internet.

    1. A ship headed due east is moving through the water at a constant speed of 8 miles per hour. However, the true course of the ship is 60. If the currents are a constant 4 miles per hour, what is the ground speed of the ship? (Round your answer to the nearest whole number.)
    http://i40.tinypic.com/dgpkr4.png

    2. A plane headed due east is traveling with an airspeed of 290 miles per hour. The wind currents are moving with constant speed in the direction 240. If the ground speed of the plane is 145 miles per hour, what is its true course?
    http://i40.tinypic.com/2vng9z6.png

    Help would be appreciated SO much!
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  2. #2
    MHF Contributor

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    Re: Law of Sine the Ambiguous Case Problems

    Draw a horizontal line, of length 8, representing the heading of the ship. Draw a second line, no set length, at 30 degrees above the first line ("due east" is 90 degrees from North so 60 degrees is 90- 60= 30 degrees from due east), representing the true path. Using compasses, strike a circle with center at the tip of the horizontal line of radius 4, representing the current of unknown direction. That circle will strike the line at 30 degrees
    1) not at all, if the radius is not large enough
    2) exactly once if the radius is exactly enough to just touch the line
    3) twice if the radius is greater than that distance (this is the "ambiguous case".)

    The "ground speed" of the ship is the length of the third side of that triangle.
    Use the "cosine law". It says that if a triangle has sides of length a, b, and c with angles A, B, and C, each opposite the corresponding lower letter, then
    c^2= a^2+ b^2- 2ab cos(C).

    Since the only angle we know is the 30 degree angle, lets use that as "C". Then c= 4, a= 8 and b is unknown:
    16= 64+ b^2- 16b cos(30)
    16= 64+ b^2- 8\sqrt{3}b
    b^2- 8\sqrt{3}b+ 48= 0
    A quadratic equation may have 0, 1, or 2 real roots corresponding to the three geometric cases.

    Go through that and then try the second problem yourself.
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