document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=890000>Question 890000</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In triangle ABC, angle A is 23 degrees, c=29cm, and a=12.2cm. How do I draw two different triangles with these properties and find the measure of angle C in each case?\r<BR>\n" );
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document.write( "I've attempted to draw two different triangles but I don't understand, won't I end up with the same angle C in both triangles if they have these same properties? \r<BR>\n" );
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document.write( "Thank you for any help you can give!\r<BR>\n" );
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document.write( "Hanna </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #538648 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=538648\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> \r<BR>\n" );
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document.write( "see this link that discusses the ambiguous case, which is what you have.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.regentsprep.org/Regents/math/algtrig/ATT12/lawofsinesAmbiguous.htm\" target = \"_blank\">http://www.regentsprep.org/Regents/math/algtrig/ATT12/lawofsinesAmbiguous.htm</a>\r<BR>\n" );
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document.write( "here's another link from the same website that also discusses the same thing.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.regentsprep.org/Regents/math/geometry/GP4/Ltriangles.htm\" target = \"_blank\">http://www.regentsprep.org/Regents/math/geometry/GP4/Ltriangles.htm</a>\r<BR>\n" );
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document.write( "here's a picture of your two triangles.<BR>\n" );
document.write( "see below the picture for further comments.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2014/072902.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "angle A is given as equal to 23 degrees and can't change.<BR>\n" );
document.write( "side a is opposite angle A and is given as equal to 12.2 and can't change.<BR>\n" );
document.write( "side c is opposite angle C and is given as equal to 29 and can't change.\r<BR>\n" );
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document.write( "side b is not given and can change.<BR>\n" );
document.write( "angle B and angle C are not given and can change.\r<BR>\n" );
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document.write( "you can make two triangldes as shown in the diagram.\r<BR>\n" );
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document.write( "this is because angle C has two possibilities.<BR>\n" );
document.write( "the first possibility is an acute angle.<BR>\n" );
document.write( "the second possibility is an obtuse angle.<BR>\n" );
document.write( "angle B is determined by the size of angle C.<BR>\n" );
document.write( "in both cases, the sum of the angles is equal to 180, so the two triangles are possible.\r<BR>\n" );
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document.write( "use the law of sines to find angle C.\r<BR>\n" );
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document.write( "a/sin(A) = c/sin(C) becomes 12.2/sin(23) = 29/sin(C)<BR>\n" );
document.write( "solve for sin(C) to get:<BR>\n" );
document.write( "sin(C) = 29*sin(23)/12.2 = .92878...<BR>\n" );
document.write( "angle C is equal to arcsin(.92878...) = 68.2465... degrees which i will round off to 68 degrees for display purposes.<BR>\n" );
document.write( "this makes angle B equal to 180 - 23 - 68.2465... = 88.75347... which i will round off to 89 degrees for display purposes.\r<BR>\n" );
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document.write( "use the law of sines to find side b.\r<BR>\n" );
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document.write( "a/sin(A) = b/sin(B) becomes 12.2 / sin(23) = b/sin(88.75347...<BR>\n" );
document.write( "solve for b to get:<BR>\n" );
document.write( "b = 12.2 * sin(88.75347...) / sin(23) = 31.2161... which i'll round off to 31.2 for display purposes.\r<BR>\n" );
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document.write( "angle C is an acute angle and can possibly be an obtuse angle as well.<BR>\n" );
document.write( "the equivalent angle in quadrant 2 is equal to 180 - C which is equal to 180 - 68.2465... which becomes equal to 111.7534... which i'll round off to 112 for display purposes.  That becomes angle C'\r<BR>\n" );
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document.write( "since the sum of the angles of a triangle is always 180 degrees, that makes angle B' = 180 - A - C' = 45.2465... degrees which i'll round off to 45 degrees for display purposes.\r<BR>\n" );
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document.write( "use the law of sines again to find the length of b' \r<BR>\n" );
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document.write( "a/sin(A) = b'/sin(B') becomes:<BR>\n" );
document.write( "12.2/sin(23) = b'/sin(45.2465...)<BR>\n" );
document.write( "solve for b' to get:<BR>\n" );
document.write( "b' = 12.2 * sin(45.2465...) / sin(23) = 22.1731... which i'll round off to 22.2 for display purposes.\r<BR>\n" );
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document.write( "you have two triangle, each of which has side c = 29, side a = 12.2, and angle A = 23 degrees.\r<BR>\n" );
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document.write( "what has changed are side b at 31.2 to side b' at 22.2, angle C at 68 degrees to angle C' at 112 degrees, and angle B at 89 degrees to angle B' at 45 degrees, as shown on the display.\r<BR>\n" );
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document.write( "both triangle are possible, given that angle A is 23 degrees and side c is 29 units and side a is 12.2 units.\r<BR>\n" );
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document.write( "if two triangles were not possible, more then likely the law of sines would have given you a value for sine that is not possible, such as something greater than 1.\r<BR>\n" );
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document.write( "you can also draw an altitude from vertex B to side b and find the measure of that altitude.\r<BR>\n" );
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document.write( "if side a is smaller than the altitude than 0 triangles are possible.<BR>\n" );
document.write( "if side a is equal to that altitude, than 1 triangle is possible.<BR>\n" );
document.write( "if side a is greater than that altitude, than two triangles are possible.\r<BR>\n" );
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document.write( "in all the cases above, side a would have to be less than side c.<BR>\n" );
document.write( "if side a is greater than side c, then only one triangle is possible.\r<BR>\n" );
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document.write( "in your triangle, the alritude came out to be 11.3 or something like that.<BR>\n" );
document.write( "side a was greater than that and less than side c so two triangles were possible.\r<BR>\n" );
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document.write( "best thing to do if you don't know is to test it out and see if it makes sense.\r<BR>\n" );
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document.write( "the following video give you a practial test to see if you can make zero triangles, one triangle, or two triangles.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.youtube.com/watch?v=Io3xZMOrbkQ\" traget = \"_blank\">http://www.youtube.com/watch?v=Io3xZMOrbkQ</a>\r<BR>\n" );
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document.write( " </SPAN>\n" );
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