SOLUTION: Sir/Madam;
A pleasant day! Kindly help me with this word problem about Law of Cosines. here it is:
Two ships leave harbor at 9:00 AM in the directions S 72°W and S 25°E
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-> SOLUTION: Sir/Madam;
A pleasant day! Kindly help me with this word problem about Law of Cosines. here it is:
Two ships leave harbor at 9:00 AM in the directions S 72°W and S 25°E
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Question 826255: Sir/Madam;
A pleasant day! Kindly help me with this word problem about Law of Cosines. here it is:
Two ships leave harbor at 9:00 AM in the directions S 72°W and S 25°E at 27 kph and 35 kph respectively. How far apart are they at 9:55 AM?
Thank you very much for your help! Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! Note that both angles are measured from the y axis south of the x axis. 72 degrees south west is measured clockwise from the y axis and 25 degrees south east is measured from the y axis counter clockwise.
We use rate * time = distance to calculate two sides of our triangle and note that the third side is the distance between our two ships.
Note that both ships are sailing for 55 minutes and the rates are given in kilometers per hour.
side 1 = 27 * 55/60 = 24.75
side 2 = 35 * 55/60 = 32.08
also add 72 and 25 to get the angle between side 1 and side 2 and
we have side angle side (SAS) information about the triangle
we can now use the law of cosines
c^2 = a^2 +b^2 - 2ab * cosine(C)
c^2 = (24.75)^2 +(32.08)^2 - 2*(24.75)*(32.08)*cosine(97)
c^2 = 612.56 + 1029.13 - 1587.96*(-0.12)
c^2 = 1832.25
c = 42.80
our two ships are 42.8 kilometers apart after sailing for 55 minutes
