Question

A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 27 mi/hr. Another ship leaves the same port at 1:30 P.M. and travels S20°W at 18 mi/hr. Approximately how far apart are the ships at 3:00 P.M.? (Round your answer to the nearest mile.)

Answer 1

To solve this problem you must apply the proccedure shown below:

1. You must apply the Law of Cosines, as you can see in the figure attached. Then:

- The first ship travels at
`27 mi/hr` in for two hours. Therefore, the side
`a` is:

`a=(27 mi/hr)(2 hr)=54mi`

- The second ship travels at
`18 mi/hr` for
`1.5 hours`. Therefore, the side
`b` is:

`b=(18mi/hr)(1.5hr)=27mi`

- Now, you can calculate
`c`:

`c=\sqrt{54^(2)+27^(2)-2(54)(27)Cos(55)}=44 mi`

The answer is:
`44 miles`