document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=886161>Question 886161</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The measure of &#8736;A is 25° more than three times the measure of &#8736;B. The sum of their measures is 165°. Find <BR>\n" );
document.write( "m&#8201;&#8736;A<BR>\n" );
document.write( " and <BR>\n" );
document.write( "m&#8201;&#8736;B. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #535741 by </B><A HREF=/tutors/aboutme.mpl?userid=basketball1127><B>basketball1127(19)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=basketball1127><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=535741\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = measure of &#8736;A\r<BR>\n" );
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document.write( "Given that,<BR>\n" );
document.write( "&#8736;A = 3&#8736;B + 25° and<BR>\n" );
document.write( "&#8736;A + &#8736;B = 165°\r<BR>\n" );
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document.write( "x = 3&#8736;B + 25°<BR>\n" );
document.write( "x + &#8736;B = 165°\r<BR>\n" );
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document.write( "Substituting,<BR>\n" );
document.write( "(3&#8736;B+25°)+&#8736;B = 165°<BR>\n" );
document.write( "4&#8736;B+25° = 165°<BR>\n" );
document.write( "4&#8736;B = 165°-25°<BR>\n" );
document.write( "4&#8736;B = 140°<BR>\n" );
document.write( "&#8736;B = 35°\r<BR>\n" );
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document.write( "&#8736;A + &#8736;B = 165°<BR>\n" );
document.write( "&#8736;A + 35° = 165°<BR>\n" );
document.write( "&#8736;A = 165°-35°<BR>\n" );
document.write( "&#8736;A =130° </SPAN>\n" );
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