document.write( "Question 865281: Find the polar coordinates of the point whose rectangular coordinates are (4sqrt 3, -4)\r
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Algebra.Com's Answer #521635 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
Find the polar coordinates of the point whose rectangular coordinates are
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document.write( "We will indicate the rectangular coordinates of point P in \r\n" );
document.write( "black as \"P%28x%2Cy%29\" and the polar coordinates of point P\r\n" );
document.write( "in red as \"red%28P%28r%2Ctheta%29%29\"\r\n" );
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document.write( "Plot P using its rectangular coordinates \"P%28x%2Cy%29\" = (\"P%284sqrt%283%29%2C0%29\":\r\n" );
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document.write( "The x-coordinate of point P is \"x=4sqrt%283%29\" and the y-coordinate\r\n" );
document.write( "of point P is -4.\r\n" );
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document.write( "Draw a right triangle with this point P and the origin as\r\n" );
document.write( "vertices and the right angle on the x-axis.  The legs of this right\r\n" );
document.write( "triangle x and y are the RECTANGULAR coordinates of point P. The \r\n" );
document.write( "hypotenuse r of this right triangle is the first POLAR coordinate \r\n" );
document.write( "of P.  The angle \"theta\" indicated by the counter-clockwise red \r\n" );
document.write( "arc is the second POLAR coordinate of the point \"red%28P%28r%2Ctheta%29%29\" \r\n" );
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document.write( "We only need to calculate \"red%28r%29\" and \"+theta\"\r\n" );
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document.write( "\"red%28r%29%5E2=x%5E2%2By%5E2\"\r\n" );
document.write( "\"red%28r%29%5E2=%284sqrt%283%29%29%5E2%2B%28-4%29%5E2\"\r\n" );
document.write( "\"red%28r%29%5E2=16%2A3%2B16\"\r\n" );
document.write( "\"red%28r%29%5E2=48%2B16\"\r\n" );
document.write( "\"red%28r%29%5E2=64\"\r\n" );
document.write( "\"red%28r%29=8\"\r\n" );
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document.write( "\"tan%28red%28theta%29%29=y%2Fx=%28-4%29%2F%284sqrt%283%29%29=-1%2Fsqrt%283%29\"\r\n" );
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document.write( "Therefore \"red%28theta%29\" in the 4th quadrant is \"red%2811pi%2F6%29\", and\r\n" );
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document.write( "\"P%284sqrt%283%29%2C0%29\" = \"red%28P%288%2C11pi%2F6%29%29\"\r\n" );
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\n" ); document.write( "Find the rectangular coordinates of the point whose polar coordinates are
\n" ); document.write( "\"red%28P%28-4%2Cpi%2F6%29%29\".
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document.write( "Let's draw the point \"red%28P%28-4%2Cpi%2F6%29%29\" using its polar coordinates.\r\n" );
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document.write( "First we draw the angle \"red%28theta%29\" with a dotted line through\r\n" );
document.write( "the origin:\r\n" );
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document.write( "Next we locate the value of r on the x-axis, then we swing an\r\n" );
document.write( "arc from that point on the x-axis around to the dotted line,\r\n" );
document.write( "like the green arc below swinging from -4 on the x-axis to\r\n" );
document.write( "the dotted line, and mar that point \"red%28P%28-4%2Cpi%2F6%29%29\".\r\n" );
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document.write( "Then we erase the green arc and draw a right triangle with \r\n" );
document.write( "this point P and the origin as vertices and the right angle \r\n" );
document.write( "on the x-axis, and indicate the RECTANGULAR coordinates x,\r\n" );
document.write( "y, of the point P.  Since we swung the point from the point\r\n" );
document.write( "x=-4, the hypotenuse \"red%28r=-4%29\".\r\n" );
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document.write( "Now we calculate x and y.\r\n" );
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document.write( "\"cos%28theta%29=x%2Fred%28r%29\"\r\n" );
document.write( "\"cos%28pi%2F6%29=x%2Fred%28-4%29\"\r\n" );
document.write( "\"sqrt%283%29%2F2=x%2Fred%28-4%29\"\r\n" );
document.write( "\"2x=red%28-4%29sqrt%283%29\"\r\n" );
document.write( "\"x=-2sqrt%283%29\"\r\n" );
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document.write( "\"sin%28theta%29=y%2Fred%28r%29\"\r\n" );
document.write( "\"sin%28pi%2F6%29=y%2Fred%28-4%29\"\r\n" );
document.write( "\"y=red%28-4%29sin%28pi%2F6%29\"\r\n" );
document.write( "\"y=red%28-4%29%281%2F2%29\"\r\n" );
document.write( "\"y=-2\"\r\n" );
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document.write( "So \"red%28P%28-4%2Cpi%2F6%29%29\" = \"P%28-2sqrt%283%29%2C-2%29\"\r\n" );
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\n" ); document.write( "Find a rectangular form of the equation \"red%28r+=+5+cos%28theta%29%29\"
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document.write( "Always substitute trig functions first, \r\n" );
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document.write( "\"sin%28red%28theta%29%29=y%2Fred%28r%29\", \"cos%28red%28theta%29%29=x%2Fred%28r%29\", \"tan%28red%28theta%29%29=y%2Fx\"\r\n" );
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document.write( "and always wait until last to replace \"red%28r%29\" by \"red%28r%29=sqrt%28x%5E2%2By%5E2%29\"\r\n" );
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document.write( "\"red%28r%29=+5cos%28theta%29\"\r\n" );
document.write( "\"red%28r%29+=+5%28x%2Fred%28r%29%29\"\r\n" );
document.write( "\"red%28r%29+=+5x%2Fred%28r%29\"\r\n" );
document.write( "\"red%28r%5E2%29+=+5x\"\r\n" );
document.write( "\"%28sqrt%28x%5E2%2By%5E2%29%29%5E2+=+5x\"\r\n" );
document.write( "\"x%5E2%2By%5E2=5x\"\r\n" );
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document.write( "You can then put it in standard form of a circle,\r\n" );
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document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\"\r\n" );
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document.write( "getting:\r\n" );
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document.write( "\"%28x-5%2F2%29%5E2%2B%28y-0%29%5E2=%285%2F2%29%5E2\"\r\n" );
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document.write( "whose graph is a circle with center (h,k) = (\"5%2F2\",0) \r\n" );
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document.write( "and radius \"r=5%2F2\": \r\n" );
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document.write( "Edwin
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