document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1201106>Question 1201106</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A vertical aerial AB 9.6m high stands on ground which is inclined 12 degrees to the horizontal. A stay connects the top of the aerial A to a point C on the ground 10m downhill from B the foot of the aerial. Determine the length of the stay and the angle the stay makes with the ground. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #853590 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=853590\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> AB =9.6 m<BR>\n" );
document.write( "%28alpha%29 = 12 deg. %28beta%29+=+%2890-%28alpha%29%29<BR>\n" );
document.write( "90-12 = 78<BR>\n" );
document.write( "Law of cosines<BR>\n" );
document.write( "AC^2 =AB^2+BC^2 −2(AB)(BC)cos∠ABC<BR>\n" );
document.write( "= (9.6)^2 + (10)^2 - 2*(10)*(9.6)*cos (78deg)<BR>\n" );
document.write( "=152.24 m<BR>\n" );
document.write( "length of the stay = 152.24 m<BR>\n" );
document.write( "Law of sines<BR>\n" );
document.write( "(sin∠ACB/AB ) = (sin∠ABC/AC)<BR>\n" );
document.write( "Find the angle  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );