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7 October, 12:28

A ship leaves port on a bearing of 34.0 and travels 10.4 mi. the ship then turns due east and travels 4.6mi. how far is the ship from port, and what is its bearing from port?

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  1. 7 October, 14:57
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    A bearing of 34° corresponds to corresponding angle of θ=90-34=56°

    The (x, y) values for the position of the ship after completing its first heading are:

    x = (10.4cos 56)

    y = (10.4sin56)

    The trigonometric angle for θ=90-90=0

    The (x, y) values for the postion of the ship after completing the second bearing is:

    x = (10.4 cos56) + (4.6cos0) ≈10.4 mi

    y = (10.4sin56) + (4.6sin0) ≈8.6mi

    the distance from the port will therefore be:

    d=√ (10.4²+8.6²) ≈13.5 miles

    It trigonometric angles is:

    θ=arctan (y/x)

    θ=arctan (8.6/10.4)

    θ≈39.6°

    Thus the bearing angle is:

    90-39.6=50.4°
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