document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1159690>Question 1159690</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 6. 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?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #782741 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=782741\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Diagram<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"https://i.imgur.com/DetohqE.png\"><BR>\n" );
document.write( "A = ship's starting point and the location of the Columbia Bar<BR>\n" );
document.write( "B = ship's location 2 hours after leaving point A<BR>\n" );
document.write( "C = ship's location 3 hours after leaving point B<BR>\n" );
document.write( "D = point used to set up bearing of S30W\r<BR>\n" );
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document.write( "1 knot = 1 nautical mile per hour<BR>\n" );
document.write( "15 knots = 15 nautical miles per hour<BR>\n" );
document.write( "After 2 hours, the ship travels a distance of 2*15 = 30 nautical miles. So AB = 30.<BR>\n" );
document.write( "Similarly, BC = 45 since 3*15 = 45\r<BR>\n" );
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document.write( "Angle DAB = 30 degrees. We start by facing directly south, then turn 30 degrees westward, which is what the notation \"S30W\" indicates. \r<BR>\n" );
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document.write( "For more information and 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( "For now, ignore point D. Focus solely on triangle ABC.<BR>\n" );
document.write( "Triangle ABC is a right triangle with the 90 degree angle at point B. This is due to the fact that the ship turns 90 degrees to the southeast.\r<BR>\n" );
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document.write( "tan(angle) = opposite/adjacent<BR>\n" );
document.write( "tan(A) = BC/AB<BR>\n" );
document.write( "tan(A) = 45/30<BR>\n" );
document.write( "tan(A) = 1.5<BR>\n" );
document.write( "A = arctan(1.5)<BR>\n" );
document.write( "A = 56.3099324740202  You need to make sure your calculator is in degree mode<BR>\n" );
document.write( "So angle BAC is roughly 56.3099324740202 degrees\r<BR>\n" );
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document.write( "Subtract off the green angle 30 degrees to get <BR>\n" );
document.write( "(angle BAC) - (angle DAB) = 56.3099324740202 - 30 = 26.3099324740202<BR>\n" );
document.write( "which is the approximate measure of the red angle DAC.\r<BR>\n" );
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document.write( "If we're situated at point A, face directly south, and turn roughly 26.3099324740202 degrees eastward, then we'll be looking directly at point C. \r<BR>\n" );
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document.write( "Rounding to 4 decimal places, that would be roughly 26.3099; feel free to change the level of accuracy based off what your teacher instructs. \r<BR>\n" );
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document.write( "The bearing to the ship from the Columbia Bar is <font color=red size=4>approximately S 26.3099 E</font><BR>\n" );
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