document.write( "Question 934939: A smuggler leaves a private airfield at 06:00 and flies on a course of
\n" ); document.write( "N40degreesE at 200 km/h.
\n" ); document.write( "The plane is detected by radar at the police airport, which
\n" ); document.write( "is located 150 km northwest of the airfield. At 06:30 the police airplane leaves its airport with the intention of intercepting the smuggler at 08:30. Determine the course the police airplane must travel, to the nearest unit. \r
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Algebra.Com's Answer #568195 by vicgonzerx(31)\"\" \"About 
You can put this solution on YOUR website!
LET A=SMUGGLER STARTING POINT @ 6:OO
\n" ); document.write( "B=SMUGGLER POINT @ 6:30
\n" ); document.write( "C= POLICE POINT WHEN DETECTING THE SMUGGLER @6:30
\n" ); document.write( "D= POINT WHERE THE SMUGGLER INTERCEPTED BY THE POLICE @ TIME 8:30
\n" ); document.write( "rEQUIRED IS THE COURSE OF THE POLICE DURING INTERCEPTION==? \r
\n" ); document.write( "\n" ); document.write( "@ 6:30 THE SMUGGLER TRAVELED THE DISTANCE OF AB @ RATE 200KPH TIME TRAVELED IS 0.5HR\r
\n" ); document.write( "\n" ); document.write( "AB=VT
\n" ); document.write( "AB=200*.5
\n" ); document.write( "AB=100KM
\n" ); document.write( "@ TIME 8:30 THE SMUGGLER TRAVELED AT DISTANCE BD @ THE RATE OF 200KPH
\n" ); document.write( "SO..BD=200(8.5-6.5)=400KM\r
\n" ); document.write( "\n" ); document.write( "SEE ATTACHED DIAGRAM FOR THE BASIS OF GETTING THIS COURSE
\n" ); document.write( "403.89N 18.46DEG E------------ THE ANSWER \r
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