document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=980796>Question 980796</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm having difficulties solving this problem. Could someone please help me?\r<BR>\n" );
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document.write( "Two forces with magnitudes of 131 and 52 newtons act on point with a resultant force of 100 <BR>\n" );
document.write( "newtons. Find the angle, in radians, between the forces.\r<BR>\n" );
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document.write( "Enter answer accurate to four decimal places.\r<BR>\n" );
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document.write( "Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #601927 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=601927\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> While I was busy drawing pretty pictures, Tim was posting a more succinct answer.<BR>\n" );
document.write( "I am posting my answer too, hoping that my drawings, and my different angle (pun intended) help too.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C100%2C-45%2C135%2C-15%2C45%2C%0D%0Atriangle%280%2C0%2C-37.65%2C35.865%2C93.35%2C35.865%29%2C%0D%0Atriangle%28131%2C0%2C0%2C0%2C93.35%2C35.865%29%2C%0D%0Agreen%28arrow%280%2C0%2C131%2C0%29%29%2Cgreen%28arrow%280%2C0%2C-37.65%2C35.865%29%29%2C%0D%0Ared%28arrow%280%2C0%2C93.35%2C35.865%29%29%2Clocate%2846.7%2C17.9%2Cred%28100%29%29%2C%0D%0Alocate%2865%2C0%2Cgreen%28131%29%29%2Clocate%2829%2C35.8%2C131%29%2C%0D%0Alocate%28-27%2C20%2Cgreen%2852%29%29%2Clocate%28103%2C22%2C52%29%2C%0D%0Agreen%28arc%280%2C0%2C40%2C40%2C-136.4%2C0%29%29%2Clocate%280%2C19%2Cgreen%28theta%29%29%2C%0D%0Aarc%28131%2C0%2C40%2C40%2C-180%2C-136.4%29%2Clocate%28114%2C10%2CA%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > I am assuming that you have to find the angle <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=green%28theta%29 ALIGN=MIDDLE  ALT=\"green%28theta%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "but we can also find the other angle in the parallelogram,<BR>\n" );
document.write( "which is supplementary to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=green%28theta%29 ALIGN=MIDDLE  ALT=\"green%28theta%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > : <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=pi-green%28theta%29 ALIGN=MIDDLE  ALT=\"A=pi-green%28theta%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "The resultant divides that parallelogram into two (congruent) triangles.<BR>\n" );
document.write( "We need to apply to one of those triangles the law of cosines,<BR>\n" );
document.write( "which is sort-of like an extension of the Pythagorean theorem,<BR>\n" );
document.write( "for triangles that are not necessarily right triangles.<BR>\n" );
document.write( "For <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C100%2C-45%2C135%2C-15%2C45%2C%0D%0Atriangle%28131%2C0%2C0%2C0%2C93.35%2C35.865%29%2C%0D%0Agreen%28line%280%2C0%2C131%2C0%29%29%2C%0D%0Ared%28line%280%2C0%2C93.35%2C35.865%29%29%2Clocate%2846.7%2C20%2Cred%28a%29%29%2C%0D%0Alocate%2865%2C0%2Cgreen%28b%29%29%2C%0D%0Alocate%28109%2C28%2Cc%29%2C%0D%0Aarc%28131%2C0%2C40%2C40%2C-180%2C-136.4%29%2Clocate%28115%2C10%2CA%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > law of cosines says <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2=b%5E2%2Bc%5E2-2%2Ab%2Ac%2Acos%28A%29 ALIGN=MIDDLE  ALT=\"a%5E2=b%5E2%2Bc%5E2-2%2Ab%2Ac%2Acos%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "Applying law of cosines to the triangle above, with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=system%28a=100%2Cb=131%2Cc=52%2CA=pi-green%28theta%29%29 ALIGN=MIDDLE  ALT=\"system%28a=100%2Cb=131%2Cc=52%2CA=pi-green%28theta%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > , we get<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=100%5E2=131%5E2%2B52%5E2-2%2A131%2A52%2Acos%28A%29 ALIGN=MIDDLE  ALT=\"100%5E2=131%5E2%2B52%5E2-2%2A131%2A52%2Acos%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and form that we get<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28A%29=%28131%5E2%2B52%5E2-100%5E2%29%2F%282%2A131%2A52%29=0.724090 ALIGN=MIDDLE  ALT=\"cos%28A%29=%28131%5E2%2B52%5E2-100%5E2%29%2F%282%2A131%2A52%29=0.724090\"   DISABLED_style= \"max-width:90%; height:auto;\" > (rounded)<BR>\n" );
document.write( "Since they are supplementary angles <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28green%28theta%29%29=-cos%28A%29=-0.724090 ALIGN=MIDDLE  ALT=\"cos%28green%28theta%29%29=-cos%28A%29=-0.724090\"   DISABLED_style= \"max-width:90%; height:auto;\" > (rounded) .<BR>\n" );
document.write( "Inverse cosine function tells us that the angles, in radians are<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=green%28theta%29=2.3805 ALIGN=MIDDLE  ALT=\"green%28theta%29=2.3805\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=0.7611 ALIGN=MIDDLE  ALT=\"A=0.7611\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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