document.write( "Question 143302: The vertices of parallelogram ABCD are A(0,0), B(7,0), (12,8), and D(5,8). Find, to the nearest degree the measure of angle DAB.\r
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Algebra.Com's Answer #104299 by Edwin McCravy(20056) $\"\"$ $\"About$

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The vertices of parallelogram ABCD are
\n" ); document.write( "A(0,0), B(7,0), (12,8), and D(5,8). Find,
\n" ); document.write( "to the nearest degree the measure of angle DAB.
\n" ); document.write( "I don't get this at all! All I know is that the
\n" ); document.write( "measure of side AB is 7 and the measure of side CD is also 7.
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document.write( "First draw a perpendicular DE to AB\r\n" );
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document.write( "Because D's coordinates are (5,8), we know that AE=5, DE=8\r\n" );
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document.write( "ADE is a right triangle.\r\n" );
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document.write( "DE is the side opposite angle DAB and\r\n" );
document.write( "AE is the side adjacent angle DAB,\r\n" );
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document.write( "and since \r\n" );
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document.write( "Now, with calculator in DEGREE mode, use the \r\n" );
document.write( "key and get 57.99461679° which to the nearest degree is 58°\r\n" );
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document.write( "Edwin

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