document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1139071>Question 1139071</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man walks a distance of 10km from his house to the church on a bearing of 56 degrees . He then walked to the market a distance of 5km on a bearing of 146 degrees .Calculate the bearing of the house from the market.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #756854 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=756854\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> my worksheet is shown below:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2019/041413.jpg\" alt=\"$$$\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the bearing of 56 degrees is angle BAE.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "bearings are the angles made from the north line going clockwise.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "line AD and line EB are parallel and congruent.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "line EA and BD are parallel and congruent.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this forms rectangle EBDA.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "line AD is the diagonal to this rectangle.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it is the distance from point A to point B.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "point A is the house.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "point B is the church.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "once at the church, the market is 5 km away at a bearing of 146 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that will be point C.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "angle FBC is the angle that represents bearing of 146 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "angle FBC is composed of angle FBG which is 90 degrees, and angle GBC which is 56 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since angle GBD is 90 degrees, angle CBD is equal to 34 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since AB is a diagonal of rectangle EBDA, then angle EAB is equal to angle DBA, each being 56 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "what you have is:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the house at point A and the church at point B and the market at point C form triangle ABC.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this triangle has a 90 degree angle at point B which is angle ABC.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that makes triangle ABC a right triangle with leg AB equal to 10 kilometers and leg BC equal to 5 kilometers.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the angle you want is angle BAC which is labeled as angle A1.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since triangle ABC is a right triangle, then tan(A1) = opp/adj = 5/10.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "arctan(5/10) is equal to 26.56505118 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the bearing from point A to point C is therefore 56 + 26.56505118 degrees = 82.56505118 degrees.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that should be your solution.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the bearing is the addition of angle EAB and angle BAC.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if angle ABC was, for some reason, not equal to 90 degrees, then you could still have found angle BAC through use of the law of cosines.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "any questions, email dtheophilis@gmail.com.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "note that lines AE and FD are vertical.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "also note that lines EG and HD are horizontal.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );