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

Answer 1

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).

Answer 2

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