Question 531343: Two ships leave port at the same time. One ship takes a bearing of 26± East of due North. The other ship takes a bearing of 32± West of due North. Both ships are travelling at a speed of 30 miles per hour. How far apart, to the nearest mile, will they be at the end of three hours?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Two ships leave port at the same time. One ship takes a bearing of 26± East of due North. The other ship takes a bearing of 32± West of due North. Both ships are travelling at a speed of 30 miles per hour. How far apart, to the nearest mile, will they be at the end of three hours?
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let x= distance ships will be apart at the end of 3 hours
You are working with a triangle where the included angle=26+32=58º
At 30 mph after 3 hours each of the two sides=90 mi
So, Law of Cosines can be used to solve for x.
..
x^2=90^2+90^2-2*90*90*cos58º
x^2=16200-8584.69=7615.31
x=√(7615.31)=87.27 mi
ans:
Distance ships will be apart at the end of 3 hours=87.27 mi
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