document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=250081>Question 250081</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An aircraft is vertically above a point which is 10km West and 15km North of a control tower.  If the aircraft is 4000m above the ground, how far is it from the control tower? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #185307 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=185307\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We are dealing with three dimensional space and we need to first calculate the ground distance from the point on the below the aircraft to the tower (which is the hypotenuse of the ground triangle with legs 10 and 15) This is also the other leg of the triangle with one leg 4  and the distance from the aircraft to the tower is the hypotenuse<BR>\n" );
document.write( "Let's calculate the first hypotenuse.<BR>\n" );
document.write( "10^2+15^2=c^2=325<BR>\n" );
document.write( "now we have the two legs squared of the other triangle.<BR>\n" );
document.write( "16 and 325<BR>\n" );
document.write( "16+325=c^2=341<BR>\n" );
document.write( "c=18.47 km from control tower to the aircraft.<BR>\n" );
document.write( "Tutor Alan combined it all.\r<BR>\n" );
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