Two ships leave a port at 9 A.M. One travels at a bearing of N 53 degrees W at 12 mph and the other travels at a bearing of S 67 degrees W at 16 mph. Approximate how far apart they are at noon that day.

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- Jul 12th 2009, 06:00 PMjordangiscoolLaw of Cosine Word Problems Pt. 2
Two ships leave a port at 9 A.M. One travels at a bearing of N 53 degrees W at 12 mph and the other travels at a bearing of S 67 degrees W at 16 mph. Approximate how far apart they are at noon that day.

- Jul 12th 2009, 08:47 PMSoroban
Hello, jordangiscool!

Did you make a sketch?

Quote:

Two ships leave a port at 9 A.M.

One travels at a bearing of N 53° W at 12 mph

and the other travels at a bearing of S 67° W at 16 mph.

Approximate how far apart they are at noon that day.

Code:`A * N`

/ * |

/ 36 *53°|

/ * |

/ 60° * P

/ * |

/ 48 * 67° |

/ * |

/ * S

/ *

B *

The first ship leaves the port $\displaystyle P$, where $\displaystyle \angle NPA = 63^o$

. . and sails 36 miles to point $\displaystyle A.$

The second ship leaves the port $\displaystyle P$, where $\displaystyle \angle SPB = 67^o$

. . and sails 48 miles to point $\displaystyle B.$

We see that $\displaystyle \angle APB = 60^o.$

Law of Cosines: .$\displaystyle AB^2 \:=\:36^2 + 48^2 - 2(36)(48)\cos60^o$

Got it?