document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197251>Question 1197251</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the interior angles of the triangle whose vertices are (-3,3), (-1,-1), (3,-3) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #830489 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=830489\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "A = (-3,3)<BR>\n" );
document.write( "B = (-1,-1)<BR>\n" );
document.write( "C = (3,-3)\r<BR>\n" );
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document.write( "Use the distance formula to calculate the distance from A to B, which gives the length of side AB.<BR>\n" );
document.write( "(x1,y1) = (-3,3) and (x2, y2) = (-1,-1)<BR>\n" );
document.write( "d = sqrt( (x1-x2)^2 + (y1-y2)^2 )<BR>\n" );
document.write( "d = sqrt( (-3-(-1))^2 + (3-(-1))^2 ) <BR>\n" );
document.write( "d = sqrt( (-3+1)^2 + (3+1)^2 )<BR>\n" );
document.write( "d = sqrt( (-2)^2 + (4)^2 )<BR>\n" );
document.write( "d = sqrt( 4 + 16 )<BR>\n" );
document.write( "d = sqrt( 20 )\r<BR>\n" );
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document.write( "Segment AB is exactly sqrt(20) units long.<BR>\n" );
document.write( "This is side c since angle C is opposite this side.\r<BR>\n" );
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document.write( "Repeat similar steps for BC and AC.\r<BR>\n" );
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document.write( "The sides of triangle ABC are<BR>\n" );
document.write( "a = segment BC = sqrt(20)<BR>\n" );
document.write( "b = segment AC = sqrt(72)<BR>\n" );
document.write( "c = segment AB = sqrt(20)\r<BR>\n" );
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document.write( "Now we can use the law of cosines to find angle A<BR>\n" );
document.write( "a^2 = b^2 + c^2 - 2*b*c*cos(A)<BR>\n" );
document.write( "(sqrt(20))^2 = (sqrt(72))^2 + (sqrt(20))^2 - 2*sqrt(72)*sqrt(20)*cos(A)<BR>\n" );
document.write( "20 = 72+20 - 2*sqrt(72)*sqrt(20)*cos(A)<BR>\n" );
document.write( "20 = 92 - 2*sqrt(72*20)*cos(A)<BR>\n" );
document.write( "20 - 92 = -2*sqrt(1440)*cos(A)<BR>\n" );
document.write( "-72 = -2*sqrt(144*10)*cos(A)<BR>\n" );
document.write( "-72 = -2*sqrt(144)*sqrt(10)*cos(A)<BR>\n" );
document.write( "-72 = -2*12*sqrt(10)*cos(A)<BR>\n" );
document.write( "-72 = -24*sqrt(10)*cos(A)<BR>\n" );
document.write( "cos(A) = -72/(-24*sqrt(10))<BR>\n" );
document.write( "cos(A) = 0.94868329805051<BR>\n" );
document.write( "A = arccos(0.94868329805051)<BR>\n" );
document.write( "A = 18.4349488229227<BR>\n" );
document.write( "A = 18.435\r<BR>\n" );
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document.write( "Angle A is roughly 18.435 degrees.<BR>\n" );
document.write( "Since triangle ABC is isosceles, this means angle C is also this approximate measure. <BR>\n" );
document.write( "The congruent angles are opposite the congruent sides. Use the law of cosines to confirm this statement.\r<BR>\n" );
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document.write( "The remaining angle B is,<BR>\n" );
document.write( "A+B+C = 180<BR>\n" );
document.write( "B = 180-(A+C)<BR>\n" );
document.write( "B = 180-(18.4349488229227+18.4349488229227)<BR>\n" );
document.write( "B = 143.130102354154<BR>\n" );
document.write( "B = 143.130\r<BR>\n" );
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document.write( "Summary:<BR>\n" );
document.write( "Angle A = 18.435 degrees<BR>\n" );
document.write( "Angle B = 143.130 degrees<BR>\n" );
document.write( "Angle C = 18.435 degrees<BR>\n" );
document.write( "Each value is approximate.<BR>\n" );
document.write( "Round these values according to the instructions your teacher gives you.<BR>\n" );
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