document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=824448>Question 824448</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a &#916;ABC, angle B = 90 degrees, AB = 6 units and AC = 10 units then the length of angle bisector AD from angle A is </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #497346 by </B><A HREF=/tutors/aboutme.mpl?userid=tommyt3rd><B>tommyt3rd(5050)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tommyt3rd><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=tommyt3rd><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=497346\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cosA=6/10=3/5\r<BR>\n" );
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document.write( "A=arccos(3/5) - about 53.13 deg<BR>\n" );
document.write( "so half is about A2=26.57 deg\r<BR>\n" );
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document.write( "therefore <BR>\n" );
document.write( "cosA2=6/AD<BR>\n" );
document.write( "cos26.57=6/AD<BR>\n" );
document.write( "AD=6/cos26.57=6.71 </SPAN>\n" );
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