document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=482589>Question 482589</A>: <I><SPAN STYLE=\"word-break: break-all;\"> for a triangle when c = 60 degree prove 1/b+c + 1/a+c = 3/a+b+c </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #330779 by </B><A HREF=/tutors/aboutme.mpl?userid=cleomenius><B>cleomenius(959)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=cleomenius><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=cleomenius><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=330779\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I assume you are working with an equilateral triangle.\r<BR>\n" );
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document.write( "b = 60 degrees. Given<BR>\n" );
document.write( "a = 60 degrees The angles in an equilateral tringle are equal.<BR>\n" );
document.write( "c = 60 degrees The angles in an equilateral triangle are equal.<BR>\n" );
document.write( "1/b + c = 1/120   addition<BR>\n" );
document.write( "1/a + c = 1/120   addition\r<BR>\n" );
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document.write( "1/120 + 1/120  = 2/120  = 1/60  addition, reduction.\r<BR>\n" );
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document.write( "3/a + b  + c = 3/180 substitution\r<BR>\n" );
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document.write( "3/180 = 1/60  Reduction.<BR>\n" );
document.write( "They are the same, 1/60.<BR>\n" );
document.write( "Cleomenius.\r<BR>\n" );
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