document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1180032>Question 1180032</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider the angle shown below with an initial ray pointing in the 3-o'clock direction that measures θ radians (where 0≤θ<2π. The circle's radius is 2.3 units long and the terminal point is (−1.35,1.86)\r<BR>\n" );
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document.write( "a) What is the slope of the terminal ray?<BR>\n" );
document.write( "   m=\r<BR>\n" );
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document.write( "b) Then, tan−1(m)=\r<BR>\n" );
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document.write( "c) Does the number we get in part (b) give us the correct value of θ? ? Yes No \r<BR>\n" );
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document.write( "d) Therefore, θ=<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #809618 by </B><A HREF=/tutors/aboutme.mpl?userid=MathLover1><B>MathLover1(20855)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=MathLover1><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=MathLover1><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=809618\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The circle's radius is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2.3 ALIGN=MIDDLE  ALT=\"2.3\"   DISABLED_style= \"max-width:90%; height:auto;\" > units long and the terminal point is (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.35 ALIGN=MIDDLE  ALT=\"-1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.86 ALIGN=MIDDLE  ALT=\"1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0%3C=+theta+%3C2pi ALIGN=MIDDLE  ALT=\"0%3C=+theta+%3C2pi\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "the point  (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.35 ALIGN=MIDDLE  ALT=\"-1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.86 ALIGN=MIDDLE  ALT=\"1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >) is in Q II and tan is negative\r<BR>\n" );
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document.write( "Since the position in the 3-o'clock is basically the positive x-axis, this means the angle that is terminating at (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.35 ALIGN=MIDDLE  ALT=\"1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.86 ALIGN=MIDDLE  ALT=\"-1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >) (which is in Q IV) should look like: \r<BR>\n" );
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document.write( "sketch\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+drawing%28+600%2C+600%2C+-5%2C+5%2C+-5%2C+5%2C%0D%0Acircle%280%2C0%2C2.3%29%2C+circle%28-1.35%2C1.86%2C+.1%29%2Cblue%28arrow%28-1.35%2C1.86%29%29%2Cgreen%28arrow%281.35%2C-1.86%29%29%2Clocate%28.5%2C-.5%2Ctheta=54.027612950715096%29%2Cgreen%28line%281.35%2C0%2C1.35%2C-1.86%29%29%2C%0D%0Alocate%28-1.5%2C.5%2Chighlight%28theta=-54.027612950715096%29%29%2C%0D%0Agraph%28+600%2C+600%2C+-5%2C+5%2C+-5%2C+5%2C+0%29%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > \r<BR>\n" );
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document.write( "Notice that we can draw a right triangle by drawing a vertical straight line from the point (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.35 ALIGN=MIDDLE  ALT=\"1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.86 ALIGN=MIDDLE  ALT=\"-1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >).\r<BR>\n" );
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document.write( " from the right triangle we have<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28m%29=1.86%2F-1.35=-1.3777777777777778 ALIGN=MIDDLE  ALT=\"tan%28m%29=1.86%2F-1.35=-1.3777777777777778\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "a) What is the slope of the terminal ray?\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m=-1.3777777777777778 ALIGN=MIDDLE  ALT=\"m=-1.3777777777777778\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "b) Then, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%5E-1%28m%29=tan%5E-1%28-1.3777777777777778%29=-54.027612950715096 ALIGN=MIDDLE  ALT=\"tan%5E-1%28m%29=tan%5E-1%28-1.3777777777777778%29=-54.027612950715096\"   DISABLED_style= \"max-width:90%; height:auto;\" >°\r<BR>\n" );
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document.write( "c) Does the number we get in part (b) give us the correct value of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=theta ALIGN=MIDDLE  ALT=\"theta\"   DISABLED_style= \"max-width:90%; height:auto;\" >? <BR>\n" );
document.write( " <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+No ALIGN=MIDDLE  ALT=\"+No\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "d) Therefore, since the point  (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.35 ALIGN=MIDDLE  ALT=\"-1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.86 ALIGN=MIDDLE  ALT=\"1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >) is in Q II,<BR>\n" );
document.write( "we then subtract it from <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=360 ALIGN=MIDDLE  ALT=\"360\"   DISABLED_style= \"max-width:90%; height:auto;\" >  because the coordinates of the point (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.35 ALIGN=MIDDLE  ALT=\"1.35\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1.86 ALIGN=MIDDLE  ALT=\"-1.86\"   DISABLED_style= \"max-width:90%; height:auto;\" >) are in the 4th quadrant, we need angle in Q II\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=theta=360-54.027612950715096 ALIGN=MIDDLE  ALT=\"theta=360-54.027612950715096\"   DISABLED_style= \"max-width:90%; height:auto;\" >°<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=theta=305.972387049284904 ALIGN=MIDDLE  ALT=\"theta=305.972387049284904\"   DISABLED_style= \"max-width:90%; height:auto;\" >°\r<BR>\n" );
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