document.write( "Question 667490: A triangle with < BAC = 20 DEG.and AB=AC. D IS such a point on AB that AD=BC.Then find out < ADC.
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Algebra.Com's Answer #415115 by AnlytcPhil(1810)\"\" \"About 
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document.write( "Here is the triangle without the point D.\r\n" );
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document.write( "Before we put in the point D, let's chop the isosceles \r\n" );
document.write( "triangle into two right triangles with a median to the \r\n" );
document.write( "base, like this green line AE, and we will let BE = 1 unit\r\n" );
document.write( "making BC = 2 units:\r\n" );
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document.write( "\"AB%2FBE\" = sec(80°), \"AB%2F1\" = sec(80°), AB = sec(80°) = AC\r\n" );
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document.write( "Now we have a case of Side-angle-side with triangle ADC.\r\n" );
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document.write( "So we use the law of cosines first to find the length of CD.\r\n" );
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document.write( "CD˛ = AD˛ + AC˛ - 2·AD·AC·cos(20°)\r\n" );
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document.write( "CD˛ = 2˛ + sec˛(80°) - 2·2·sec(80°)·cos(20°)\r\n" );
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document.write( "CD˛ = 4 + sec˛(80°) - 4·sec(80°)·cos(20°)\r\n" );
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document.write( "CD˛ = 15.51754097\r\n" );
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document.write( "CD = 3.939231012\r\n" );
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document.write( "Now we turn to the law of sines:\r\n" );
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document.write( "\"CD%2Fsin%28A%29\" = \"AC%2Fsin%28ADC%29\"\r\n" );
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document.write( "Cross multiply:\r\n" );
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document.write( "CD·sin(ADC) = AC·sin(A)\r\n" );
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document.write( "sin(ADC) = \"%28AC%2Asin%28A%29%29%2F%28CD%29\"\r\n" );
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document.write( "sin(ADC) = \"%28%22sec%2880%B0%29%22%2Asin%28%2220%B0%22%29%29%2F%283.939231012%29\"\r\n" );
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document.write( "sin(ADC) = 0.5\r\n" );
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document.write( "< ADC = 150°\r\n" );
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document.write( "Edwin
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