document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1120894>Question 1120894</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The figure shows a triangle STU where TU=11cm. H lies on TU such that the length of TH is 120% of the length of HU and angle SHU=90degree. Given that the area of triangle STH is 21cm^2, find angle TSU. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #736667 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=736667\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Since the measure of the base is 11 and the area is 21, the height, which is to say segment SH must be <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{42}{11}\">.\r<BR>\n" );
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document.write( "Further, since TH is 120% of HU, let the value HU be designated as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">, then we can say:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 1.2x\ +\ x\ =\ 11\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 2.2x\ =\ 11\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \text{HU}\ =\ 5\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 1.2x\ =\ \text{TH}\ =\ 6\"> \r<BR>\n" );
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document.write( "For the sake of clarity, designate angle TSH as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \alpha\"> and angle USH as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \beta\">.\r<BR>\n" );
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document.write( "By definition, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \tan\alpha\ =\ \frac{TH}{SH}\ =\ \frac{6}{\frac{42}{11}}\ =\ \frac{66}{42}\">\r<BR>\n" );
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document.write( "Likewise, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \tan\beta\ =\ \frac{HU}{SH} =\ \frac{5}{\frac{42}{11}}\ =\ \frac{55}{42}\">\r<BR>\n" );
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document.write( "The tangent of the sum of two angles is given as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \tan(a\ +\ b)\ =\ \frac{\tan{a}\ +\ \tan{b}}{1\ -\ \tan{a}\tan{b}}\">, so:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \tan(\alpha\,+\,\beta)\ =\ \frac{\frac{66}{42}\,+\,\frac{55}{42}}{1-\(\frac{66}{42}\)\(\frac{55}{42}\)}\ \approx\ -2.7235\">\r<BR>\n" );
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document.write( "Since Angle TSU is clearly the sum of angle <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \alpha\"> plus angle <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \beta\">:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \angle\text{TSU}\ =\ \tan^{-1}\ \(\tan(\alpha\,+\,\beta)\)\ \approx\ \tan^{-1}(-2.7235)\">\r<BR>\n" );
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document.write( "Your calculator will give you a negative angle as an answer.  Providing you have the calculator set to work in degrees, add 180 degrees to the calculator result to get the appropriate QII positive number result.<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
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