document.write( "Question 763795: The largest interior angle of a kite is 140 °. Side measures are 1.6 m
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Algebra.Com's Answer #465521 by Edwin McCravy(20060)\"\" \"About 
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document.write( "We want the shorter diagonal BD, \r\n" );
document.write( "but first we must calculate the longer diagonal, \r\n" );
document.write( "the green line AC below:\r\n" );
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document.write( "We use the law of cosines on ΔABC since this is the SAS case:\r\n" );
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document.write( "AC² = AB² + BC² - 2·AC·BC·cos(∠ABC)\r\n" );
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document.write( "AC² = 1.2² + 1.6² - 2(1.2)(1.6)cos(140°)\r\n" );
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document.write( " AC = 2.634693656 m\r\n" );
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document.write( "We use the law of sines to find ∠BAC\r\n" );
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document.write( "\"BC%2Fsin%28BAC%29\"\"%22%22=%22%22\"\"AC%2Fsin%28B%29\"\r\n" );
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document.write( "\"1.6%2Fsin%28BAC%29\"\"%22%22=%22%22\"\"2.634693656%2Fsin%28%22140%B0%22%29\"\r\n" );
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document.write( "Cross multiply:\r\n" );
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document.write( "(2.634693656)sin(∠BAC) = (1.6)sin(140°)\r\n" );
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document.write( "             sin(∠BAC) = \"%281.6sin%28%22140%B0%22%29%29%2F2.634693656\"\r\n" );
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document.write( "                  ∠BAC = 22.97645649°\r\n" );
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document.write( "Now that we have ∠BAC we draw half the shorter diagonal BE, \r\n" );
document.write( "which is perpendicular to the longer diagonal AC, in red:\r\n" );
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document.write( "Since ΔABE is a right triangle:\r\n" );
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document.write( "\"%28BE%29%2FAB\" = \"%28opposite%29%2F%28hypotenuse%29\" = sin(BAC)\r\n" );
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document.write( "BE = AB·sin(BAC)\r\n" );
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document.write( "BE = 1.2·sin(22.97645649°)\r\n" );
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document.write( "BE = 0.4684234191 m\r\n" );
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document.write( "So the entire shorter diagonal, BD, \r\n" );
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document.write( "is twice the length of BE, so\r\n" );
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document.write( "BD = 2·BE = 2(0.4684234191) = 0.9368468381 m \r\n" );
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document.write( "Rounded to tenths:  0.9 m\r\n" );
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document.write( "Edwin

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