Expert: Richard J. Raridon - 3/31/2005

Question
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Followup To
Question -
1) a ship proceeds on a course of 280 degrees for 3 hours at a speed of 12 knots, then changes to a course to 200 degrees for 2 more hours. how far is the ship from its starting point?
Answer -
After three hours the ship will be 6.25 nautical miles east and 35.35 nm south.  Then it will go 8.21 nm south and 22.55 nm west.
From its starting point it will be 16.30 nautical miles west and 43.56 nautical miles south


how did u get the answer(s) to that because i don't understand how u got them. can u show me how u got the answers as well?

Answer
I presume you know about sines and cosines. Think of the ship's course as a vector that you need to resolve into x and y components.  A course of 280 degrees is 10 degrees east of due south.  After 3 hours the ship has gone 36 nautical miles.  The east component is 36sin(10) = 6.25.
The south component is 36cos(10) = 35.35.  Then the ship goes on a course of 200 degrees which is 20 degrees south of west.  So for that leg, the west component is 24cos(20) which is 22.55 and the south component is 24sin(20) = 8.21.
Then you just add up the x and y components to get the overall value.  The direct distance from the starting point is (16.30^2 +43.56^2)^1/2