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.)

Respuesta :

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 [tex] 27 mi/hr [/tex] in for two hours. Therefore, the side [tex] a [/tex] is:

[tex] a=(27 mi/hr)(2 hr)=54mi [/tex]

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

[tex] b=(18mi/hr)(1.5hr)=27mi [/tex]

- Now, you can calculate [tex] c [/tex]:

[tex] c=\sqrt{54^{2}+27^{2}-2(54)(27)Cos(55)}=44 mi [/tex]

The answer is: [tex] 44 miles [/tex]

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