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alf123 · Jun 15, 2008, 01:03 PM

Two boats named Spritzer1 and Whaler2 leave the same dock at the same time. Spritzer1 travels at a speed of 15.0km/h. The Whaler 2 heads due east and travels at a speed of 18.0km/h. After 3/4 of an hour, the boats are 14.0 km apart when Spritzer 1 calls Whaler 2 for help, because their engine has stopped working.
a) Assmuing that both boats travel in a straight part, determine the angle between their paths. (round to the nearest degree).
b) Using compass directions, what course did Spritzer1 set when leaving the dock.
c) When Spritzer 1 calls Whaler 2 for help, the captain of Whaler 2 sets a new course. What angle does the captain of Whaler 2 turn the boat?

galactus · Jun 15, 2008, 01:21 PM

The boats travel for 45 minutes. That means Spritzer travels 11.25 km and Whaler travels 13.5 km. The distance between them is 14 km

So, you have a triangle with side lengths b=13.5, c=11.25, a=14

We can use the law of cosines and solve for the angle.

$a^{2}=b^{2}+c^{2}-2bc\cdot{cosA}$

Solve for A:

$A=cos^{-1}\left(\frac{a^{2}-b^{2}-c^{2}}{2bc}\right)$

Enter your givens and you have the angle you need. You can use that to solve the other portions.

For part b, you need the angle from the positive y-axis (north) to vertex A.

For part c, You need angle ABC.