document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1077637>Question 1077637</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi can someone help me with this problem please? THANK YOU \r<BR>\n" );
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document.write( "     State the exact value of t,given cot t=-root3 and t [0,2pi) terminates QII \r<BR>\n" );
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document.write( "      Select one <BR>\n" );
document.write( "              a.5pi/6<BR>\n" );
document.write( "              B.7pi/6<BR>\n" );
document.write( "              C. pi/3<BR>\n" );
document.write( "              D. 2pi/3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692107 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=692107\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You do not need to memorize anything to figure out<BR>\n" );
document.write( "the trigonometric ratios of a <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=30-60-90 ALIGN=MIDDLE  ALT=\"30-60-90\"   DISABLED_style= \"max-width:90%; height:auto;\" > right triangle<BR>\n" );
document.write( "(one with acute angles measuring <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F6=30%5Eo ALIGN=MIDDLE  ALT=\"pi%2F6=30%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F3=60%5Eo ALIGN=MIDDLE  ALT=\"pi%2F3=60%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > ).<BR>\n" );
document.write( "That triangle is half of an equilateral triangle.<BR>\n" );
document.write( "If the length of the hypotenuse is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1 ALIGN=MIDDLE  ALT=\"1\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "the short leg's length is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-1.1%2C1.1%2C-0.2%2C1.8%2C%0D%0Agreen%28triangle%280%2C0%2C1%2C0%2C0%2C1.732%29%29%2C%0D%0Agreen%28rectangle%280%2C0%2C0.05%2C0.05%29%29%2C%0D%0Atriangle%28-1%2C0%2C1%2C0%2C0%2C1.732%29%2Clocate%280.02%2C0.8%2Ch%29%2C%0D%0Alocate%280.4%2C0.1%2C%221+%2F+2%22%29%2Clocate%280.45%2C0.88%2C1%29%2C%0D%0Ared%28arc%280%2C1.732%2C0.6%2C0.6%2C60%2C90%29%29%2C%0D%0Alocate%280.05%2C1.45%2Cred%28pi%2F6%29%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "and the Pythagorean theorem lets you figure out<BR>\n" );
document.write( "the length of the other leg as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=sqrt%283%29%2F2 ALIGN=MIDDLE  ALT=\"h=sqrt%283%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "So, for the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F6=30%5Eo ALIGN=MIDDLE  ALT=\"pi%2F6=30%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > , we have the trigonometric ratios<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28pi%2F6%29=1%2F2 ALIGN=MIDDLE  ALT=\"sin%28pi%2F6%29=1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > , <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28pi%2F6%29=sqrt%283%29%2F2 ALIGN=MIDDLE  ALT=\"cos%28pi%2F6%29=sqrt%283%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > , and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28pi%2F6%29=%281%2F2%29%2F%28sqrt%283%29%2F2%29=1%2Fsqrt%283%29 ALIGN=MIDDLE  ALT=\"tan%28pi%2F6%29=%281%2F2%29%2F%28sqrt%283%29%2F2%29=1%2Fsqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "Although cotangent is not as popular as sine, cosine, and tangent,<BR>\n" );
document.write( "it is defined as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cot%28theta%29=1%2Ftan%28theta%29 ALIGN=MIDDLE  ALT=\"cot%28theta%29=1%2Ftan%28theta%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<BR>\n" );
document.write( "so <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cot%28pi%2F6%29=sqrt%283%29 ALIGN=MIDDLE  ALT=\"cot%28pi%2F6%29=sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "You know how angles in other quadrants relate to<BR>\n" );
document.write( "symmetrical acute reference angles in quadrant I.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-1.2%2C1.2%2C-1.2%2C1.2%2Cgrid%280%29%2C%0D%0Ared%28circle%280%2C0%2C1%29%29%2Carrow%280%2C0%2C1.2%2C0.8%29%2C%0D%0Aarrow%280%2C0%2C-1.2%2C0.8%29%2Carrow%280%2C0%2C-1.2%2C-0.8%29%2C%0D%0Aarrow%280%2C0%2C1.2%2C-0.8%29%2Carc%280%2C0%2C1%2C1%2C-33.7%2C0%29%2C%0D%0Alocate%280.36%2C0.2%2Ctheta%29%2Carc%280%2C0%2C1%2C1%2C180%2C213.7%29%2C%0D%0Alocate%28-0.4%2C0.2%2Ctheta%29%2Carc%280%2C0%2C1.2%2C1.2%2C0%2C33.7%29%2C%0D%0Alocate%280.45%2C-0.1%2Ctheta%29%2Carc%280%2C0%2C1.2%2C1.2%2C146.3%2C180%29%2C%0D%0Alocate%28-0.5%2C-0.1%2Ctheta%29%2Carc%280%2C0%2C1.8%2C1.8%2C180%2C360%29%2C%0D%0Alocate%28-0.63%2C0.63%2Cpi%29%2Ctriangle%28-0.9%2C0%2C-0.93%2C0.05%2C-0.87%2C0.05%29%2C%0D%0Ared%28arc%280%2C0%2C1.4%2C1.4%2C213.7%2C360%29%29%2Clocate%280.07%2C0.63%2Cred%28pi-theta%29%29%2C%0D%0Ared%28triangle%28-0.58%2C0.39%2C-0.54%2C0.39%2C-0.58%2C0.43%29%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "In Quadrant II (QII for short), we have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi-pi%2F6=5pi%2F6=5pi%2F6 ALIGN=MIDDLE  ALT=\"pi-pi%2F6=5pi%2F6=5pi%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "as the symmetrical angle to reference angle <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F6=5pi%2F6 ALIGN=MIDDLE  ALT=\"pi%2F6=5pi%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "The absolute values of the trigonometric functions for those two angles are the same,<BR>\n" );
document.write( "but some signs change:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%285pi%2F6%29=sin%28pi%2F6%29=1%2F2 ALIGN=MIDDLE  ALT=\"sin%285pi%2F6%29=sin%28pi%2F6%29=1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > , <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%285pi%2F6%29=-cos%28pi%2F6%29=-sqrt%283%29%2F2 ALIGN=MIDDLE  ALT=\"cos%285pi%2F6%29=-cos%28pi%2F6%29=-sqrt%283%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%285pi%2F6%29=-tan%28pi%2F6%29=-1%2Fsqrt%283%29 ALIGN=MIDDLE  ALT=\"tan%285pi%2F6%29=-tan%28pi%2F6%29=-1%2Fsqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > , and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cot%285pi%2F6%29=-cot%28pi%2F6%29=-sqrt%283%29 ALIGN=MIDDLE  ALT=\"cot%285pi%2F6%29=-cot%28pi%2F6%29=-sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " <BR>\n" );
document.write( "NOTE:<BR>\n" );
document.write( "To be able to say that you know trigonometry,<BR>\n" );
document.write( "you should be able to figure out (not memorize without understanding)<BR>\n" );
document.write( "the trigonometric functions for certain key angles.<BR>\n" );
document.write( "You should start with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0=0%5Eo ALIGN=MIDDLE  ALT=\"0=0%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F2=90%5Eo ALIGN=MIDDLE  ALT=\"pi%2F2=90%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "based on the unit circle definitions,<BR>\n" );
document.write( "because you cannot use right triangle trigonometric ratios for those angles.<BR>\n" );
document.write( "From there, you should also know the trigonometric functions' values<BR>\n" );
document.write( "for all angles that are multiples of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F2=90%5Eo ALIGN=MIDDLE  ALT=\"pi%2F2=90%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "Beyond that, there are key angles between <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0=0%5Eo ALIGN=MIDDLE  ALT=\"0=0%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F2=90%5Eo ALIGN=MIDDLE  ALT=\"pi%2F2=90%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "whose trigonometric ratios you should be able to figure out.<BR>\n" );
document.write( "Consider a square of side length <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1 ALIGN=MIDDLE  ALT=\"1\"   DISABLED_style= \"max-width:90%; height:auto;\" > , split in half by a diagonal.<BR>\n" );
document.write( "The two halves are isosceles right triangles,<BR>\n" );
document.write( "so both acute angles measure <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=90%5Eo%2F2=45%5Eo=pi%2F4 ALIGN=MIDDLE  ALT=\"90%5Eo%2F2=45%5Eo=pi%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "and the length, x, of the legs of that right triangle,<BR>\n" );
document.write( "can be figured out using the Pythagorean theorem.<BR>\n" );
document.write( "Since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2=x%5E2%2Bx%5E2=1%5E2=1 ALIGN=MIDDLE  ALT=\"2x%5E2=x%5E2%2Bx%5E2=1%5E2=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > , <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2=1%2F2 ALIGN=MIDDLE  ALT=\"x%5E2=1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=sqrt%281%2F2%29=1%2Fsqrt%282%29=sqrt%282%29%2F2 ALIGN=MIDDLE  ALT=\"x=sqrt%281%2F2%29=1%2Fsqrt%282%29=sqrt%282%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "No memorization needed.<BR>\n" );
document.write( "For the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=30%5Eo=pi%2F6 ALIGN=MIDDLE  ALT=\"30%5Eo=pi%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60%5Eo=pi%2F3 ALIGN=MIDDLE  ALT=\"60%5Eo=pi%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > ,<BR>\n" );
document.write( "consider an equilateral triangle with side length <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1 ALIGN=MIDDLE  ALT=\"1\"   DISABLED_style= \"max-width:90%; height:auto;\" > (as shown above)<BR>\n" );
document.write( "split into two congruent right triangles by one of its altitudes. . </SPAN>\n" );
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