document.write( "Question 543470: Two boats leave the harbour at the same time. Boat A travels at 32km/h on a bearing of 125 degrees and boat B travels at 18km/h on a bearing of 205 degrees\r
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\n" ); document.write( "\n" ); document.write( "How far apart are the boats?\r
\n" ); document.write( "\n" ); document.write( "In what direction would the skipper of boat B have to look to see boat A?
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Algebra.Com's Answer #355531 by lwsshak3(11628)\"\" \"About 
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Two boats leave the harbour at the same time. Boat A travels at 32km/h on a bearing of 125 degrees and boat B travels at 18km/h on a bearing of 205 degrees
\n" ); document.write( "After 3 hours
\n" ); document.write( "How far apart are the boats?
\n" ); document.write( "In what direction would the skipper of boat B have to look to see boat A?
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\n" ); document.write( "Draw a triangle with the apex representing the starting point where the boats left the harbor. Call this point C. From this point C, extend a 96 km line at a bearing of 125º. Call this point A.
\n" ); document.write( "Also from point C, extend a 54 km line at a bearing of 205º. Call this point B. Draw a line connecting points A and B which is the distance(c) between boats A and B. You now have a triangle with two sides, CB and CA and their included angle of 80º. Solve with law of cosines:
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\n" ); document.write( "Law of cosines: c^2=a^2+b^2-2*a*b cos80º
\n" ); document.write( "c^2=54^2+96^2-2*54*96 cos80º
\n" ); document.write( "c^2=2916+9216-10368 cos80º
\n" ); document.write( "c^2=12432-1800.38=10631.62
\n" ); document.write( "c=√10631.62≈103.11 km
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\n" ); document.write( "sin B/96=sine C/103.11=sin 80º/103.11
\n" ); document.write( "sin B=(96/103.11)*sin80º=.917
\n" ); document.write( "B≈66.5º
\n" ); document.write( "Bearing of boat A from boat B=25+66.5=91.5º
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\n" ); document.write( "ans:
\n" ); document.write( "The boats are 103.11 km apart after 3 hours.
\n" ); document.write( "The skipper of boat B must look to see boat A at a bearing of 91.5º
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\n" ); document.write( "please check my calculations
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