document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1159678>Question 1159678</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.) A DC-9 aircraft leaves Portland International Airport from runway 10 RIGHT, whose bearing is S80° E.  After flying for a half mile, the pilot requests permission to turn 90°  and head toward the southwest.  The permission is granted.  After the airplane goes 10 miles in this direction, what bearing should the control tower use to locate the aircraft? Diagram needed w/ written explanation?\r<BR>\n" );
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document.write( "2.)A ship leaves the Columbia bar with a bearing of S30° W and a speed of 15 knots.  After 2 hours, the ship turns 90° towards the southeast.  After 3 hours, maintaining the same speed, what is the bearing to the ship from the Columbia bar? Diagram needed w/ written explanation?<BR>\n" );
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document.write( "Please and thank you<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #782721 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=782721\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll do the first problem to get you started. \r<BR>\n" );
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document.write( "Diagram<BR>\n" );
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document.write( "A = plane's starting location = location of air traffic control (ATC) tower<BR>\n" );
document.write( "B = plane's location after traveling that initial 1/2 mile<BR>\n" );
document.write( "C = plane's location after turning at point B and traveling 10 miles<BR>\n" );
document.write( "angle DAB = 80 degrees, see note below<BR>\n" );
document.write( "angle GAB = 10 degrees, complementary to angle DAB<BR>\n" );
document.write( "angle EBF = 80 degrees, congruent to angle DAB<BR>\n" );
document.write( "angle EBC = 10 degrees, it is complementary to angle EBF<BR>\n" );
document.write( "angle ABC = 90 degrees, supplementary to angle CBF = 90<BR>\n" );
document.write( "angle CBF = 90 degrees, amount the pilot turns\r<BR>\n" );
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document.write( "Note: The notation \"S80E\" means you start facing south and turn 80 degrees east. For more info/examples about compass bearings, see <a  rel=nofollow HREF=\"http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/comp.html\">this page</a>.\r<BR>\n" );
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document.write( "Vector AB has length or magnitude of 0.5 miles<BR>\n" );
document.write( "It's direction is -10 degrees (-1 times the measure of angle GAB)<BR>\n" );
document.write( "So r = 0.5 and theta = -10, which means<BR>\n" );
document.write( "x = r*cos(theta)<BR>\n" );
document.write( "x = 0.5*cos(-10)<BR>\n" );
document.write( "x = 0.5*cos(10)<BR>\n" );
document.write( "and<BR>\n" );
document.write( "y = r*sin(theta)<BR>\n" );
document.write( "y = 0.5*sin(-10)<BR>\n" );
document.write( "y = -0.5*sin(10)\r<BR>\n" );
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document.write( "Vector AB is <BR>\n" );
document.write( "< x, y > = <0.5*cos(10), -0.5*sin(10)>\r<BR>\n" );
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document.write( "Vector BC has magnitude r = 10 and direction theta = -100 degrees or theta = 260 degrees depending on how you view things. So vector BC can be broken up like so<BR>\n" );
document.write( "x = r*cos(theta)<BR>\n" );
document.write( "x = 10*cos(-100)<BR>\n" );
document.write( "x = 10*cos(100)<BR>\n" );
document.write( "and<BR>\n" );
document.write( "y = r*sin(theta)<BR>\n" );
document.write( "y = 10*sin(-100)<BR>\n" );
document.write( "y = -10*sin(100)\r<BR>\n" );
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document.write( "Vector BC is <BR>\n" );
document.write( "< x, y > = <10*cos(100), -10*sin(100)>\r<BR>\n" );
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document.write( "We found<BR>\n" );
document.write( "Vector AB is < x, y > = <0.5*cos(10), -0.5*sin(10)><BR>\n" );
document.write( "Vector BC is < x, y > = <10*cos(100), -10*sin(100)>\r<BR>\n" );
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document.write( "Add up the components to get the resultant vector AC<BR>\n" );
document.write( "vector AC = < 0.5*cos(10)+10*cos(100),  -0.5*sin(10) +  -10*sin(100) ><BR>\n" );
document.write( "vector AC = < -1.2440779001632, -9.93490161895554 >\r<BR>\n" );
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document.write( "Now compute the direction for vector AC<BR>\n" );
document.write( "theta = arctan(y/x)<BR>\n" );
document.write( "theta = arctan(-9.93490161895554/(-1.2440779001632))<BR>\n" );
document.write( "theta = 82.8624052261118<BR>\n" );
document.write( "We add on 180 degrees so we move from quadrant Q1 to Q3<BR>\n" );
document.write( "theta = 82.8624052261118+180<BR>\n" );
document.write( "theta = 262.862405226111\r<BR>\n" );
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document.write( "This is fairly close to 270 degrees which is pointing directly south (when we assign 0 degrees to mean directly east)<BR>\n" );
document.write( "So, 270 - 262.862405226111 = 7.13759477388897 degrees <BR>\n" );
document.write( "is the amount the ATC has to turn when they face directly south. Doing this will have ATC align their viewpoint with the plane's new position at point C.\r<BR>\n" );
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document.write( "Therefore, the bearing ATC has to aim for is <font size=4 color=red>approximately S 7.13759 W</font><BR>\n" );
document.write( "ATC must start facing directly south, then turn approximately 7.13759 degrees toward the west.\r<BR>\n" );
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document.write( "Referring back to the diagram, angle DAC is approximately 7.13759 degrees\r<BR>\n" );
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