document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209728>Question 1209728</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$ and $\angle MHX = 48^\circ$, as shown below. Find $\angle ADX$, in degrees. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850036 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850036\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $E=(0,3)$, $F=(4,3)$, $G=(4,0)$, and $H=(0,0)$.<BR>\n" );
document.write( "Then $M = \left(\frac{0+4}{2}, \frac{3+3}{2}\right) = (2,3)$.<BR>\n" );
document.write( "$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.<BR>\n" );
document.write( "Since $MH = MX$, $X$ lies on a circle centered at $M$ with radius $\sqrt{13}$.<BR>\n" );
document.write( "Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.<BR>\n" );
document.write( "Also, $\angle MHX = 48^\circ$.<BR>\n" );
document.write( "The slope of $MH$ is $\frac{3-0}{2-0} = \frac{3}{2}$.<BR>\n" );
document.write( "Let the slope of $MX$ be $m$. Then $\tan(\angle MHX) = \tan(48^\circ) = \left|\frac{m - \frac{3}{2}}{1 + \frac{3}{2}m}\right|$.<BR>\n" );
document.write( "Since $X$ is outside the rectangle, we can assume that the angle is positive.<BR>\n" );
document.write( "We are given that $EH = 3$ and $EF = 4$. So, $E=(0,3)$, $F=(4,3)$, $G=(4,0)$ and $H=(0,0)$.<BR>\n" );
document.write( "$M$ is the midpoint of $EF$, so $M = (\frac{0+4}{2}, \frac{3+3}{2}) = (2,3)$.<BR>\n" );
document.write( "$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.<BR>\n" );
document.write( "$MX = MH = \sqrt{13}$.<BR>\n" );
document.write( "Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.<BR>\n" );
document.write( "$\vec{MH} = \langle 0-2, 0-3 \rangle = \langle -2, -3 \rangle$.<BR>\n" );
document.write( "$\vec{MX} = \langle x-2, y-3 \rangle$.<BR>\n" );
document.write( "$\cos(48^\circ) = \frac{\vec{MH} \cdot \vec{MX}}{|\vec{MH}||\vec{MX}|} = \frac{-2(x-2) - 3(y-3)}{\sqrt{13}\sqrt{13}} = \frac{-2x+4-3y+9}{13} = \frac{-2x-3y+13}{13}$.<BR>\n" );
document.write( "$13\cos(48^\circ) = -2x-3y+13$.<BR>\n" );
document.write( "$2x+3y = 13 - 13\cos(48^\circ)$.<BR>\n" );
document.write( "$A = (0,0)$, $D = (0,3)$.<BR>\n" );
document.write( "$\vec{AD} = \langle 0-0, 3-0 \rangle = \langle 0, 3 \rangle$.<BR>\n" );
document.write( "$\vec{AX} = \langle x-0, y-0 \rangle = \langle x, y \rangle$.<BR>\n" );
document.write( "$\cos(\angle DAX) = \frac{\vec{AD}\cdot\vec{AX}}{|\vec{AD}||\vec{AX}|} = \frac{3y}{3\sqrt{x^2+y^2}} = \frac{y}{\sqrt{x^2+y^2}}$.<BR>\n" );
document.write( "$\angle MHX = 48^\circ$. $\angle MHA = \arctan(\frac{2}{3}) \approx 33.69^\circ$.<BR>\n" );
document.write( "$\angle AHX = \angle MHX + \angle MHA = 48^\circ + 33.69^\circ \approx 81.69^\circ$.<BR>\n" );
document.write( "$\angle DAX = 90^\circ - \angle AHX = 90^\circ - 81.69^\circ \approx 8.31^\circ$.\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{30}$<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );