document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=549262>Question 549262</A>: <I><SPAN STYLE=\"word-break: break-all;\"> tan theta= -12/5 sin theta>0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #357628 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=357628\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A logical first thing to do in solving this problem is to determine in which of the four Cartesian quadrants the angle theta lies. Do this by examining the signs of the two trigonometric functions that you are given for theta.<BR>\n" );
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document.write( "First, notice that you are told that sin (theta) is greater than zero, meaning that sin (theta) is positive. (Think of the definition of the sine function as being the side opposite theta divided by the hypotenuse.) All trigonometric functions are positive in Quadrant I. The sine is also positive in Quadrant II where the side opposite is positive (above the x-axis), and the hypotenuse (the side that has unit length and rotates about the origin) is always considered positive regardless of the quadrant. Therefore, by being positive only in Quadrants I and II the sine (side opposite divided by hypotenuse) limits theta to Quadrants I and II and consequently theta will be between 0 and 180 degrees (equivalent to 0 to pi radians).<BR>\n" );
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document.write( "Next notice that you are told that tan (theta) is negative. (Recall that the tangent function is defined as the side opposite to the angle divided by the side adjacent to the angle.) To be negative the tangent must have the side opposite to the angle and the side adjacent to the the angle be of opposite signs. Recall that in Quadrant IV, the side opposite the angle will be negative (below the x-axis) and the side adjacent to the angle will be on the positive x-axis. Therefore, the tangent will be negative in Quadrant IV. Next think about Quadrant II. The side opposite the angle will be positive (above the x-axis) and the side adjacent to the angle will be on the negative x-axis. Therefore, the tangent will be negative in Quadrant II.<BR>\n" );
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document.write( "So, we have the positive sine value limiting the answer to Quadrants I and II and the negative tangent value limiting the answer to Quadrants II and IV. Theta must be in Quadrant II because that is the only Quadrant that meets both criteria, positive sine and negative tangent.<BR>\n" );
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document.write( "Now that we have determined the Quadrant in which theta is positioned, we can move on by working to find where the rotating hypotenuse is positioned. We have been given that:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28theta%29+=+-12%2F5 ALIGN=MIDDLE  ALT=\"tan%28theta%29+=+-12%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Dividing -12 by 5 converts the value of the tangent to -2.4 so we can change the equation for the tangent to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%28theta%29+=+-2.4 ALIGN=MIDDLE  ALT=\"tan%28theta%29+=+-2.4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "That being the case, we can use a scientific calculator to look up the angle that has -2.4 as its tangent. The method to do this will vary depending on the calculator, but what we are going to use is the function <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tan%5E-1%28x%29 ALIGN=MIDDLE  ALT=\"tan%5E-1%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > which should be read as \"the angle whose tangent is x.\" In our case, x is -2.4. So on a typical scientific calculator set for the degrees mode, you enter -2.4 and press the tan^-1 function to find that the answer is (in degrees) -67.3801. In Quadrant II this angle is measured from the negative y-axis clockwise to the hypotenuse which is the rotating line of unit length and with one of its ends at the origin.<BR>\n" );
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document.write( "But theta is measured from the positive x-axis counter-clockwise (or if you prefer anti-clockwise) to the hypotenuse. That means that theta is an obtuse angle equal to 180 degrees minus the 67.3801 degrees between the negative x axis and the hypotenuse. This results in theta being equal to 112.6199 degrees. And that's the answer you are looking for. <BR>\n" );
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document.write( "If you want, you can convert this angle to radians by solving for x in the proportion that reads 112.6199 degrees is to 360 degrees as x radians is to 2*pi radians. You should find that the answer for theta in radians is 1.9656.<BR>\n" );
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document.write( "Check by again ensuring that your calculator is in the degrees mode. Then clear and enter 112.6119 and press the tan key. Your answer should be -2.399995864 and this rounds to -2.4. This is the correct value, equivalent to -12/5. You can clear this and again enter 112.6199. This time press the sin key to find that the sine of 112.6199 degrees is +0.923076687. Although we were not given a required value for the sine function, this calculation verifies that the sine of an angle of 112.6199 degrees is positive as it was required to be.<BR>\n" );
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document.write( "Hope this helps you to see your way through this problem. The method is fundamental to understanding the relationship of trig functions to the Cartesian coordinate system, so you should get an understanding of what is involved in solving this problem because it will help you considerably in working similar problems. <BR>\n" );
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