Two cruise ships leave the same port with a 35° angle between their path. Cruise A is traveling at 18 miles per hour and Cruise B is traveli

Question

Two cruise ships leave the same port with a 35° angle between their path. Cruise A is traveling at 18 miles per hour and Cruise B is traveling at 15 miles per hour. If they travel in a straight path, find the distance between the cruise ships after 2 hours

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Thành Công 3 years 2021-08-22T04:08:03+00:00 2 Answers 221 views 0

Answers ( )

  1. thienhuong
    0
    2021-08-22T04:09:15+00:00
    Reply

    9514 1404 393

    Answer:

      20.7 miles

    Step-by-step explanation:

    The distance that A travels in the same direction as B is …

      (18 mi/h)(2 h)cos(35°) = 29.489 mi

    So, the difference in distances in that direction is …

      (15 mi/h)(2 h) -29.489 mi = 0.511 mi

    __

    The distance A travels in the direction perpendicular to B is …

      (18 mi/h)(2 h)sin(35°) = 20.649 mi

    So, the straight-line distance between the ships is the hypotenuse of the right triangle with these distances as legs:

      AB = √(0.511² +20.649²) = √426.632 . . . miles

      AB = 20.655 miles ≈ 20.7 miles . . . separation after 2 hours

  2. Doris
    1
    2021-08-22T04:09:26+00:00
    Reply

    Answer:

    20\:\mathrm{miles}

    Step-by-step explanation:

    After travelling two hours, the two cruise ships form a triangle. One of the legs  of this triangle will be the distance Cruise A travelled, and another will be the distance Cruise B travelled. We can find the distance they travel using:

    s=\frac{d}{t}, d=s\cdot t

    Cruise A is travelling at 18 miles per hour for 2 hours. Therefore, Cruise A has travelled:

    d=s\cdot t=18\cdot 2=36\:\mathrm{miles}

    Cruise B is travelling at 15 miles per hour for 2 hours. Therefore, Cruise B has travelled:

    d=s\cdot t=15\cdot 2=30\:\mathrm{miles}

    Because we are given the angle between these legs, we can use the Law of Cosines to find the third leg. The Law of Cosines is given by:

    c^2=a^2+b^2-2ab\cos C, where a, b, and c are interchangeable. Let c represent the distance between the two cruises. We have:

    c^2=36^2+30^2-2\cdot36\cdot30\cdot \cos 35^{\circ},\\c^2=426.63,\\c\approx \fbox{$20\:\mathrm{miles}$}(one significant figure).

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