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Two ships leave the same port at 10:00 A.M. Ship A travels at the constant rate of 25 MPH at a bearing of S70E; Ship B travels at a constant rate of 20 MPH at a bearing of N29W. How far apart are the ships at noon?
\n" ); document.write( "b. what is the bearing to Ship B from Ship A at noon?\r
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\n" ); document.write( "\n" ); document.write( "At noon ship A would have traveled 2*25=50 miles.
\n" ); document.write( "At noon ship B would have traveled 2*20=40 miles.\r
\n" ); document.write( "\n" ); document.write( "Draw an (x,y) coordinate graph with center where both ships left at 10 AM from
\n" ); document.write( "the same port.Ship A with a bearing of S70E makes a 20 degree angle with the x-axis. Ship B with a bearing of N29W makes a 29 degree angle with the y-axis.
\n" ); document.write( "You now have a triangle with sides of 50 mi and 40 mi with their included angle = 20+90+29=139 degrees.
\n" ); document.write( "Using Law of Cosines: c^2=a^2+b^2-2abCosx=50^2+40^2-2*50*40*cos 139 deg.
\n" ); document.write( "c^2=4100-(-3018.34)=7118.34
\n" ); document.write( "c=84.37 miles
\n" ); document.write( "ans:
\n" ); document.write( "The ships are 84.37 miles apart at noon.\r
\n" ); document.write( "\n" ); document.write( "..
\n" ); document.write( "bearing to ship B from ship A
\n" ); document.write( "sinx/sin40=sin139/84.37
\n" ); document.write( "sinx=40sin139/84.37=.311
\n" ); document.write( "x=18.12 degrees
\n" ); document.write( "bearing angle=90-20-18.12=51.88 deg
\n" ); document.write( "ans:
\n" ); document.write( "After 12 noon,bearing to ship B from ship A=N51.88W\r
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