document.write( "Question 1120425: A fishing boat leaves port at 10 miles per hour at a bearing of 280∘ for 5 hours, then turns to a bearing of 60∘ at 6 miles per hour for 4 hours, and finally changes to a bearing of 130∘ at 10 miles per hour for 4 hours. At this point, the boat heads directly back to port at a speed of 5 miles per hour.
\n" ); document.write( "Find the time it takes the boat to return to port as well as the boat's bearing as it does. \n" ); document.write( "

Algebra.Com's Answer #737527 by Boreal(15235) $\"\"$ $\"About$

You can put this solution on YOUR website!
One can draw this in parts.
\n" ); document.write( "Going at 280 degrees, 10 degrees north of west, the boat is 50*sin 10 north (8.68 miles) and 50 cos 10 west (west 49.24 miles)
\n" ); document.write( "Then at a bearing of 60 degrees for 24 miles is 12 degrees further north and 12 sqrt(3) miles east
\n" ); document.write( "Then at a bearing of 130 degrees for 40 miles is 40 sin 40 south or 25.71 miles south and 40 cos 40 east or 30.64 miles east.
\n" ); document.write( "These add up to -5.03 miles (or south) and -2.18 miles (east)
\n" ); document.write( "The boat has to go back 5.03 miles north and 2.18 miles west
\n" ); document.write( "The tangent is the ratio of those two numbers or 2.31, and tan(-1) of 2.31 is 66.6 degrees, and that is north of west, so the bearing is 336.6 degrees.
\n" ); document.write( "The distance is the sqrt (5.03^2+2.18^2)=5.48 miles, which is done in 1.10 hours or 1 hour6min \n" ); document.write( "

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