document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=35595>Question 35595</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If anglesA,B, and C are the sides of a triangle such that sin(A+B)=1/cos(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #22513 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=22513\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If anglesA,B, and C are the sides of a triangle such that sin(A+B)=1/cos(C)<BR>\n" );
document.write( "IN A TRIANGLE A+B+C=180.HENCE <BR>\n" );
document.write( "SIN(A+B)=SIN(180-C)=SIN(C)=1/COS(C)<BR>\n" );
document.write( "SIN(C)COS(C)=1<BR>\n" );
document.write( "SIN(2C)/2=1...<BR>\n" );
document.write( "SIN?(2C)=2...IMPOSSIBLE.<BR>\n" );
document.write( "IT IS WRONG .SIN(2C) CAN NOT EQUAL 2.IT COULD BE SIN(A+B)=1+COS(C)<BR>\n" );
document.write( "THEN AS PROVED ABOVE <BR>\n" );
document.write( "SIN(C)=1+COS(C) WHICH CAN BE SOLVED TO GET C=90.PLEASE CHECK UP AND INFORM\r<BR>\n" );
document.write( "\n" );
document.write( " and cos(A+B)=cos(C),<BR>\n" );
document.write( "COS(180-C)= - COS(C)=COS(C)<BR>\n" );
document.write( "2COS(C)=0<BR>\n" );
document.write( "COS(C)=0<BR>\n" );
document.write( "C=90 DEG. HENCE ABC IS RIGHT ANGLED AT C.<BR>\n" );
document.write( " then show that the triangle is a right triangle </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );