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Math Help - Trignometry (sin law & cos law)

  1. #1
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    Trignometry (sin law & cos law)

    Three ships, A, B and C are sailing along the same straight line on a course of 240 degrees. Ship A is at the front, and ship C brings up the rear. From ship A, the bearing of navigation buoy is 025 degrees. The buoy is 2.5 km from ship A and 2 km from ships B and C. To the nearest degree what is the bearing of the buoy from ship C? from ship B?
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  2. #2
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    On sailing ships, the bearing is measured from the North axis, clockwise.
    So a bearing of 240 degrees is a ray in the 3rd quadrant that is 240 degrees clockwise from the North axis.
    It is actually, S 60deg W.
    It is actually at 210 degrees by normal, counterclockwise, measuring of angles.

    If you can draw correctly the figure, you'd have 3 ships A, B, C on a straight line that is 30deg above the x-axis or East axis.

    With A on the (0,0), the buoy is located at 2.5km, N 25deg E.
    Ship B is somewhere that is N 60deg E of A.
    Ship C is a lot fartheraway, that is also at N 60deg E of A.

    Let point P be the location of the buoy.

    In triangle APB,
    angle PAB = 90 -25 -30 = 35deg ....say, angle A.
    AP = 2.5km ------given.
    BP = 2km -------given.
    By Law Of Sines,
    2.5/sinB = 2/sinA
    Cross multiply,
    2sinB = 2.5sinA
    sinB = (2.5sinA)/2 = 2.5sin(35deg) / 2 = 0.71697 .....positive sine value.
    That means angle B is in the 1st or 2nd quadrant.

    If angle B is in the1st quadrant,
    angle B = arcsin(0.71697) = 45.8 deg or 46deg
    So, angle PBA = 46deg
    But that is not possible according to the figure.
    So we use the other angle B that is in the 2nd quadrant,
    angle B = arcsin(0.71697) = 180 -46 = 134deg --------------**

    Then, the bearing of point P from point B:
    240 +134 = 374deg
    That is more than 360....more by 14 degrees.
    Hence, the bearing of the buoy from the ship B is 014 degrees. -----answer.

    -----------------------
    For the bearing of the buoy from ship C.

    Per givens, triangle APC is like triangle APB, in that
    angle A = 35deg
    AP = 2.5km
    CP = 2km

    By Law of Sines,
    2.5/sinC = 2/sin(35deg)
    sinC = 2.5sin(35deg) / 2 = 0.71697 .......positive, same as above.

    In the 1st quadrant,
    angle C = 46deg --------okay, according to the figure.

    So, bearing of point P from point C,
    240 +46 = 286 degrees

    Hence, the bearing of the buoy from ship C is 286 degrees. ----answer.
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  3. #3
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    thanks a lot.
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