document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=955362>Question 955362</A>: <I><SPAN STYLE=\"word-break: break-all;\"> AEC,DEB and AB are straight lines.the straights lines DA and CBF are parallel.angle ABF=90°,angle CEB=55°and angle ABE=25°.find angle ADE and find angle EAB ,explaining with reason\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #583584 by </B><A HREF=/tutors/aboutme.mpl?userid=macston><B>macston(5194)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=macston><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=macston><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=583584\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Angle CEB and angle AEB lie on line AEC and since angle CEB=55 degrees, AEB=125 degrees. FBE and CBE are supplementary along line FBC.Angle FBE=angle FBA (90 degrees)+ angle ABE (25 degrees)=115 degrees, so angle CBE=65 degrees. Angle ADE is alternate to angle CBE, so angle ADE=65 degrees. In triangle ABE we have angle ABE=25 degrees, Angle AEB=125 degrees, so angle EAB=180-125-25=30 degrees<BR>\n" );
document.write( "ANSWERS: ANGLE ADE=65 degrees and angle EAB=30 degrees.  </SPAN>\n" );
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