document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=148656>Question 148656</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find all the angles in a 5-12-13 triangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #109063 by </B><A HREF=/tutors/aboutme.mpl?userid=nabla><B>nabla(475)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nabla><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=109063\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We first note that\r<BR>\n" );
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document.write( "5^2+12^2=13^2<BR>\n" );
document.write( "25+144=169<BR>\n" );
document.write( "169=169.\r<BR>\n" );
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document.write( "So, this is a right triangle by the Pythagorean Theorem. It follows that one angle is 90 degrees.\r<BR>\n" );
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document.write( "So, we need only find the measures of angles opposite 5 and 12. Let's call them a and b. Note that a+b=90.\r<BR>\n" );
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document.write( "Cos(a)=adj/hyp=12/13<BR>\n" );
document.write( "Cos(b)=adj/hyp=5/13\r<BR>\n" );
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document.write( "So we want to take ArcCos of 12/13 and 5/13 to find the measures of the remaining angles.\r<BR>\n" );
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document.write( "ArcCos(12/13)=.3948 radians<BR>\n" );
document.write( "ArcCos(5/13)=1.1760 radians\r<BR>\n" );
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document.write( "Multiply radians by 180/pi to get degrees.\r<BR>\n" );
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document.write( "So we have 90, 22.62, 67.38 measures in degrees for angles.<BR>\n" );
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