document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140526>Question 1140526</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man start from P and walk 8km on a bearing of 70 degree. He then walk 12km on the bearing of 105 degree to Q, what is the bearing of Q from P </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #761060 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=761060\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw this.<BR>\n" );
document.write( "We know two side lengths and the angle between, which is 145 degrees.  Call that side c and use Law of Cosines<BR>\n" );
document.write( "c^2=a2+b^2-2abcos C<BR>\n" );
document.write( "=64+144-2*8*12(-0.819)<BR>\n" );
document.write( "=208+157.24<BR>\n" );
document.write( "=365.24<BR>\n" );
document.write( "c=sqrt(365.24)=19.11 km<BR>\n" );
document.write( "Now use Law of Sines<BR>\n" );
document.write( "19.11/sin 145=12/sin A, where A is the angle between the original direction walked and the final bearing.<BR>\n" );
document.write( "33.32=12/sin A\r<BR>\n" );
document.write( "\n" );
document.write( "sin A=12/33.32=0.3601<BR>\n" );
document.write( "sin^-1 of 0.3601=21.11  degrees.\r<BR>\n" );
document.write( "\n" );
document.write( "That is measured from the 70 degree bearing first started, so the final bearing is 91.11 deg </SPAN>\n" );
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