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Trig

Two ships leave a harbor at the same time. One ship travels on a bearing S11 degrees W at 12 mph. The second ship travels on a bearing N75 degrees E at 9 mph. How far apart will the ships be after 3 hours?

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Erin

  1. Ship #1: 12mi/h[11o] W. of S. = 12mi/h[191o] CW,
    Ship #2: 9mi/h[75o] E. of N. = 9mi/h[75o] CW.

    d = 36mi[191o] - 27mi[75o],
    X = 36*sin191 - 27*sin75 = -32.9 mi,
    Y = 36*Cos191 - 27*Cos75 = -42.33 mi,
    d^2 =(-39.9)^2 + ( - 42.33)^2 = 3383.84,
    d = 58.2 miles apart.

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    Henry

  2. My answer is this :
    The V1 represent 12mph[S11W], and V2 represent 9mph [N75E]
    Then we can write A and B as follows:

    From the definition of velocity we have,

    V1 = d1/t
    V2 = d2/t
    d1 = V1 * t
    d2 = V2 * t
    After 3hours, the ship 1 and ship 2 will travel,

    d1 = 12mph[S11W] * 3h = 36m [S11W]
    d2 = 9mph[N75E] *3h= 27m [N75E]

    Now the two displacements, which are vectors, can be rewrite in the following x and y coordinate forms:

    d1 = (36m * cos259, 36sin259)
    d2 = (27m * cos15, 27 * sin15)

    Now, the distance between them now can be easily calculated using a distance formula,

    d = sqrt((x2-x1)^2 + (y2 - y1)^2))
    =sqrt((36cos259 -27cos15)^2 + (36sin259 -27sin15)^2)
    =53. 63947715 m

    Thus, the two ship will be apart 54m after three hours.

    V1 = (12mph* cos 15, 12mph * sin15)
    V2 = (9mph * cos259, 9mph * sin 259)

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    Daniel

  4. I made a sketch saw that I had a triangle with sides 36 and 27 and a contained angle
    of 116°
    So, using the cosine law: let the distance between them be x
    x^2 = 36^2 + 27^2 - 2(36)(27)cos116
    = 2877.19...
    x = appr 53.64 miles, which agrees with Daniel's answer.

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    Reiny

  5. Correction: Replace 39.9 with 32.9:
    d^2 = (-32.9)^2 + (-42.33)^2 = 2874.24,
    d = 53.61 miles which agrees with Daniel and Reiny's answer.

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    Henry

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