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: $44 miles$