document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=23293>Question 23293</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Point P is the intersection of the terminal arm of angle Q in standard position and the unit circle with centre (0,0).  If P is in quadrant 3 and cosQ = m, determine the coordinates of P in terms of m.\r<BR>\n" );
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document.write( "A.  (-m, sqrt(1-m^2))<BR>\n" );
document.write( "A.  (-m, -sqrt(1-m^2))<BR>\n" );
document.write( "A.  (m, sqrt(1-m^2))<BR>\n" );
document.write( "A.  (m, -sqrt(1-m^2)) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #12094 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=12094\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw the picture.<BR>\n" );
document.write( "The point is \"m\" to the left of the y-axis.<BR>\n" );
document.write( "Now cosQ =m =m/1.  The point is \"1\" away <BR>\n" );
document.write( "from the origin because it is on a unit-<BR>\n" );
document.write( "circle.  Using Pythagoras  you get <BR>\n" );
document.write( "1^2 = m^2 + ?^2.<BR>\n" );
document.write( "The ? = sqrt(1-m^2) but it is negative because<BR>\n" );
document.write( "it is a Y-value in the III quadrant.<BR>\n" );
document.write( "P is the point (m,-sqrt(1+m^2).\r<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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