document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=281002>Question 281002</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sketch the triangle and use the law of cosines to find the indicated value.round your answer ro the nearest hundreth.\r<BR>\n" );
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document.write( "22)find the angle opposite the longest side in a triangle with sides of 43 ft ,48ft, and 65ft. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #204171 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=204171\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sketch the triangle and use the law of cosines to find the indicated value.round your answer ro the nearest hundreth. <BR>\n" );
document.write( "22)find the angle opposite the longest side in a triangle with sides of 43 ft ,48ft, and 65ft.<BR>\n" );
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document.write( "Since a^2 = b^2 + c^2 -2bc*cos(A)<BR>\n" );
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document.write( "cos(A) = (b^2+c^2-a^2)/2bc<BR>\n" );
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document.write( "Your Problem:<BR>\n" );
document.write( "cos(A) = (43^2+48^2-65^2)/(2*43*48)<BR>\n" );
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document.write( "cos(A) = -0.0174<BR>\n" );
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document.write( "A = cos^-1(-0.0174) = 91 degrees<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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