document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=174457>Question 174457</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 42. use half angle formula to find cos sin and tan of 67 degrees and 30' </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #129451 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=129451\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> use half angle formula to find cos sin and tan of 67 degrees and 30'<BR>\n" );
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document.write( "sin(67 30') = sin(135/2) <BR>\n" );
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document.write( "Formula: sin(135/2) = sqrt[(1-cos(135))/2] <BR>\n" );
document.write( "= sqrt[(1-(1/sqrt(2))/2] = sqrt[0.85355..]= 0.92388<BR>\n" );
document.write( "====================================================<BR>\n" );
document.write( "cos(67 30') = cos(135/2)<BR>\n" );
document.write( "-------<BR>\n" );
document.write( "Formula: cos(135/2) = sqrt[(1+cos(135))/2] <BR>\n" );
document.write( "= sqrt[0.146447] = 0.38268..<BR>\n" );
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document.write( "tan(67 30') = tan(135/2)<BR>\n" );
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document.write( "Formula: tan(135/2) = (1-cos(135))/2 = 0.85355...<BR>\n" );
document.write( "====================================================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( " </SPAN>\n" );
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