document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185891>Question 1185891</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A ship at A is to sail to C, 56 km north and 258 km east of A. After sailing<BR>\n" );
document.write( "N25o10’E for 120 mi to P, the ship is headed toward C. Find the distance of P<BR>\n" );
document.write( "from C and the required course to reach C. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849512 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849512\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this navigation problem:\r<BR>\n" );
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document.write( "**1. Convert the initial bearing and distance to coordinates:**\r<BR>\n" );
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document.write( "*   The ship sails N25°10'E for 120 nautical miles (we'll assume the problem uses nautical miles).<BR>\n" );
document.write( "*   Convert the bearing to a decimal degree: 25° + (10'/60) = 25.167°<BR>\n" );
document.write( "*   Calculate the change in latitude (north/south): 120 * cos(25.167°) ≈ 108.64 NM<BR>\n" );
document.write( "*   Calculate the change in longitude (east/west): 120 * sin(25.167°) ≈ 50.84 NM\r<BR>\n" );
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document.write( "**2. Determine the coordinates of point P:**\r<BR>\n" );
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document.write( "*   Point A is our origin (0, 0).<BR>\n" );
document.write( "*   Point P's coordinates are (50.84, 108.64) relative to A.\r<BR>\n" );
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document.write( "**3. Determine the coordinates of point C relative to A:**\r<BR>\n" );
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document.write( "*   Point C is 258 km east and 56 km north of A.  We need to convert these to nautical miles.  1 nautical mile is approximately 1.852 km.<BR>\n" );
document.write( "*   East: 258 km / 1.852 km/NM ≈ 139.31 NM<BR>\n" );
document.write( "*   North: 56 km / 1.852 km/NM ≈ 30.24 NM<BR>\n" );
document.write( "*   Point C's coordinates are (139.31, 30.24) relative to A.\r<BR>\n" );
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document.write( "**4. Determine the coordinates of point C relative to point P:**\r<BR>\n" );
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document.write( "*   Subtract the coordinates of P from the coordinates of C:<BR>\n" );
document.write( "    *   East: 139.31 - 50.84 ≈ 88.47 NM<BR>\n" );
document.write( "    *   North: 30.24 - 108.64 ≈ -78.4 NM\r<BR>\n" );
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document.write( "**5. Calculate the distance from P to C:**\r<BR>\n" );
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document.write( "*   Use the Pythagorean theorem: distance = sqrt(east² + north²)<BR>\n" );
document.write( "*   distance = sqrt(88.47² + (-78.4)²) ≈ sqrt(7827.84 + 6146.56) ≈ sqrt(13974.4) ≈ 118.21 NM\r<BR>\n" );
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document.write( "**6. Calculate the bearing from P to C:**\r<BR>\n" );
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document.write( "*   Use the arctangent function to find the angle: angle = arctan(east / north)<BR>\n" );
document.write( "*   angle = arctan(88.47 / -78.4) ≈ -48.42°\r<BR>\n" );
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document.write( "*Since the change in North is negative and change in East is positive, the bearing will be in the fourth quadrant.*\r<BR>\n" );
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document.write( "*   To convert to a standard bearing, we measure clockwise from North. Since we are in the fourth quadrant, we will use 360 - 48.42 to get the bearing.\r<BR>\n" );
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document.write( "*   Bearing = 360 - 48.42 ≈ 311.58 or N 48.42° W\r<BR>\n" );
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document.write( "**Answers:**\r<BR>\n" );
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document.write( "*   Distance from P to C: Approximately 118.21 nautical miles.<BR>\n" );
document.write( "*   Course to reach C from P: Approximately N 48.42° W or 311.58°<BR>\n" );
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