document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=141962>Question 141962</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the length of the legs of a 45°-45°-90° triangle with a hypotenuse of 4 times the square root of 2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #103387 by </B><A HREF=/tutors/aboutme.mpl?userid=jojo14344><B>jojo14344(1513)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jojo14344><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=103387\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> well, I guess since only the hypotenuse is given we have to use trigonometric function to solve the other 2 sides (opposite and adjacent side). And mind you they'll have both same measurements since it's 45deg-45deg to a right triangle. We'll find out;<BR>\n" );
document.write( "First, we use the function \"sine\" to get one side, = opp/hyp= \"a\"/hyp<BR>\n" );
document.write( "sin45=a/4*sqrt2<BR>\n" );
document.write( "a=(sin45)(4*sq rt2), measurement of one leg<BR>\n" );
document.write( "For the other leg we use \"cosine\" because we deal the adjacent side this time= adj/hyp= \"b\"/hyp<BR>\n" );
document.write( "cos45=b/4*sq rt2<BR>\n" );
document.write( "b=(cos45)(4*sq rt2), measurement of the other leg.<BR>\n" );
document.write( "Why they're the same?<BR>\n" );
document.write( "Because sin45=cos45, check it out.<BR>\n" );
document.write( "why i use sine or cosine? review your trigo functions i guess in dealing with those legs- opposite and adjacent side.<BR>\n" );
document.write( "Hope this help.<BR>\n" );
document.write( "Thank you,<BR>\n" );
document.write( "Jojo </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );