document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=438239>Question 438239</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help with this question,I'm really stock.I need this tomorrow.Thank you..<BR>\n" );
document.write( "A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3.5km on a bearing of 047degrees temperature to point B. From B they turn and run for 2.8km on a bearing of 342degrees temperature to point C. They then run back to point A.<BR>\n" );
document.write( "(a.) Determine the distance from C back to A.<BR>\n" );
document.write( "(b.) At what bearing does the course follow from C back to A? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #303135 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=303135\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3.5km on a bearing of 047degrees temperature to point B. From B they turn and run for 2.8km on a bearing of 342degrees temperature to point C. They then run back to point A.<BR>\n" );
document.write( "(a.) Determine the distance from C back to A.<BR>\n" );
document.write( "(b.) At what bearing does the course follow from C back to A?<BR>\n" );
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document.write( "These degrees have nothing to do with temperature ???  They're compass headings.  No wonder you're \"stock.\"<BR>\n" );
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document.write( "0 degrees is North.  047 degs is 47 degrees to the East of North.  If you make a sketch, you'll see that angle CBA = 115 degrees (not a heading)<BR>\n" );
document.write( "Now you have a triangle with 2 sides and the included angle.<BR>\n" );
document.write( "Use the Cosine Law to find side AC, side b.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2+=+a%5E2+%2B+c%5E2+-+2%2Aa%2Ac%2Acos%28115%29 ALIGN=MIDDLE  ALT=\"b%5E2+=+a%5E2+%2B+c%5E2+-+2%2Aa%2Ac%2Acos%28115%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2+=+12.25+%2B+7.84+-+2%2A3.5%2A2.8%2Acos%28115%29 ALIGN=MIDDLE  ALT=\"b%5E2+=+12.25+%2B+7.84+-+2%2A3.5%2A2.8%2Acos%28115%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2+=+28.3733 ALIGN=MIDDLE  ALT=\"b%5E2+=+28.3733\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "b =~ 5.32666 km<BR>\n" );
document.write( "= distance from C to A<BR>\n" );
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document.write( "Find angle C using the Law of Sines:<BR>\n" );
document.write( "b/sin(115) = 3.5/sin(C)<BR>\n" );
document.write( "sin(C) = 3.5*sin(115)/b =~ 0.5955<BR>\n" );
document.write( "Angle C = 36.55 degs (not a heading)<BR>\n" );
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document.write( "The heading from A to C = 047 - 36.55 =~ 010<BR>\n" );
document.write( "From C to A is the reciprocal, <BR>\n" );
document.write( "Heading = 190 degrees<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );