document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #351438 by lwsshak3(11628)\"\" \"About 
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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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\n" ); document.write( "let x= distance ships will be apart at the end of 3 hours
\n" ); document.write( "You are working with a triangle where the included angle=26+32=58º
\n" ); document.write( "At 30 mph after 3 hours each of the two sides=90 mi
\n" ); document.write( "So, Law of Cosines can be used to solve for x.
\n" ); document.write( "..
\n" ); document.write( "x^2=90^2+90^2-2*90*90*cos58º
\n" ); document.write( "x^2=16200-8584.69=7615.31
\n" ); document.write( "x=√(7615.31)=87.27 mi
\n" ); document.write( "ans:
\n" ); document.write( "Distance ships will be apart at the end of 3 hours=87.27 mi \n" ); document.write( "
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