document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=826149>Question 826149</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. In triangle ABC which is right angled at B sin A = A/A+B\r<BR>\n" );
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document.write( "a) is it reasonable to state that the hypotenuse measures ( A+B) units, justify<BR>\n" );
document.write( "b) what is the value of 1) cos A , 2) tan A 3) cot C<BR>\n" );
document.write( "Thanks so much in advance:) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #497944 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=497944\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) Since sin is opposite over adjacent one would think that if the sin was <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%2F%28A%2BB%29 ALIGN=MIDDLE  ALT=\"A%2F%28A%2BB%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, then the hypotenuse would be A+B. But it might not be. The fraction <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%2F%28A%2BB%29 ALIGN=MIDDLE  ALT=\"A%2F%28A%2BB%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > might be the reduced form of the opposite/hypotenuse ratio. For example, the opposite side and the hypotenuse might be 2A and 2(A+B) or 14A and 14(A+B), etc.<br><BR>\n" );
document.write( "b) To find cos(A) use<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+1+-+sin%5E2%28A%29 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+1+-+sin%5E2%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+1+-+%28A%2F%28A%2BB%29%29%5E2 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+1+-+%28A%2F%28A%2BB%29%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+1+-+A%5E2%2F%28A%2BB%29%5E2 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+1+-+A%5E2%2F%28A%2BB%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+%28A%2BB%29%5E2%2F%28A%2BB%29%5E2+-+A%5E2%2F%28A%2BB%29%5E2 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+%28A%2BB%29%5E2%2F%28A%2BB%29%5E2+-+A%5E2%2F%28A%2BB%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+%28A%5E2%2B2AB%2BB%5E2%29%2F%28A%2BB%29%5E2+-+A%5E2%2F%28A%2BB%29%5E2 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+%28A%5E2%2B2AB%2BB%5E2%29%2F%28A%2BB%29%5E2+-+A%5E2%2F%28A%2BB%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%5E2%28A%29+=+%282AB%2BB%5E2%29%2F%28A%2BB%29%5E2 ALIGN=MIDDLE  ALT=\"cos%5E2%28A%29+=+%282AB%2BB%5E2%29%2F%28A%2BB%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28cos%5E2%28A%29%29+=+sqrt%28%282AB%2BB%5E2%29%2F%28A%2BB%29%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28cos%5E2%28A%29%29+=+sqrt%28%282AB%2BB%5E2%29%2F%28A%2BB%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "We will ignore the negative square root because all the trig functions are positive for any angle that is in a triangle (i.e. an angle from 0 to 90).<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28A%29+=+sqrt%28%282AB%2BB%5E2%29%29%2Fsqrt%28%28A%2BB%29%5E2%29 ALIGN=MIDDLE  ALT=\"cos%28A%29+=+sqrt%28%282AB%2BB%5E2%29%29%2Fsqrt%28%28A%2BB%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28A%29+=+sqrt%28%282AB%2BB%5E2%29%29%2F%28A%2BB%29 ALIGN=MIDDLE  ALT=\"cos%28A%29+=+sqrt%28%282AB%2BB%5E2%29%29%2F%28A%2BB%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "For tan(A) use sin(A)/cos(A):<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28A%29+=+%28A%2F%28A%2BB%29%29%2F%28sqrt%282AB%2BB%5E2%29%2F%28A%2BB%29%29 ALIGN=MIDDLE  ALT=\"tan%28A%29+=+%28A%2F%28A%2BB%29%29%2F%28sqrt%282AB%2BB%5E2%29%2F%28A%2BB%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The two \"little\" denominators cancel:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28A%29+=+A%2Fsqrt%282AB%2BB%5E2%29 ALIGN=MIDDLE  ALT=\"tan%28A%29+=+A%2Fsqrt%282AB%2BB%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I'll leave it up to you to rationalize the denominator.<br><BR>\n" );
document.write( "Since B is the right angle, then A and C are the two acute angles in the triangle. The two acute triangles of a right triangle always add up to 90 degrees. IOW, they are always complementary. We will use this fact for cot(C). Since that tan and cot of complimentary angles are always equal:<BR>\n" );
document.write( "cot(C) = tan(A)<BR>\n" );
document.write( "So cot(C) will have the same answer as tan(A) (above). </SPAN>\n" );
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