document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=248840>Question 248840</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if i have a right triangle with side lengths 12 and 6, what would the degree's on the inside be? (this is using sine, consine, and tangent laws) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #181313 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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=181313\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> if i have a right triangle with side lengths 12 and 6, what would the degree's on the inside be? (this is using sine, consine, and tangent laws)\r<BR>\n" );
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document.write( "Since the triangle is a right triangle the third side is the hypotenuse H and has the value sqrt(12^2 + 6^2) = sqrt(180) = 6*sqrt(5)\r<BR>\n" );
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document.write( "You can then calculate the sine (i.e. side opposite divided by hypotenuse) of the angle opposite side 6 as 6/(6*sqrt(5), the cosine of the angle opposite side 6 (that is, adjacent to side 12) as 12/(6*sqrt(5) etc.\r<BR>\n" );
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document.write( "Similarly you can find the corresponding trig values for the other angles.   </SPAN>\n" );
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