document.write( "Question 1207589: In the diagram, ABCD is a square. Find sin angle PAQ.\r
\n" ); document.write( "\n" ); document.write( "P is on BC such that BP = 2 and PC = 3, and Q is on CD such that CQ = 2 and QD = 3. \n" ); document.write( "
Algebra.Com's Answer #845557 by ikleyn(52780)\"\" \"About 
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\n" ); document.write( "In the diagram, ABCD is a square. Find sin angle PAQ.\r
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\n" ); document.write( "\n" ); document.write( "        This problem is a plain routine and can be solved easily.\r
\n" ); document.write( "\n" ); document.write( "        There is no need to make long boring calculations.\r
\n" ); document.write( "\n" ); document.write( "        True solution is very short and straightforward.\r
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document.write( "Using Pythagoras, from triangle ABP find  AP = \"sqrt%285%5E2%2B2%5E2%29\" = \"sqrt%2829%29\".\r\n" );
document.write( "Next,  \"sin%28alpha%29\" = \"2%2Fsqrt%2829%29\",  \"cos%28alpha%29\" = \"5%2Fsqrt%2829%29\".\r\n" );
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document.write( "Using Pythagoras, from triangle ADQ find  AQ = \"sqrt%285%5E2%2B3%5E2%29\" = \"sqrt%2834%29\".\r\n" );
document.write( "Next,  \"sin%28beta%29\" = \"3%2Fsqrt%2834%29\",  \"cos%28beta%29\" = \"5%2Fsqrt%2834%29\".\r\n" );
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document.write( "OK.  Now,  ∠PAQ = 90° - \"alpha\" - \"beta\";  therefore\r\n" );
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document.write( "    sin(∠PAQ) = \"sin%2890%5Eo-alpha-beta%29\" = \"cos%28alpha%2Bbeta%29\" = \r\n" );
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document.write( "              = \"cos%28alpha%29%2Acos%28beta%29+-+sin%28alpha%29%2Asin%28beta%29\" = \"%285%2Fsqrt%2829%29%29%2A%285%2Fsqrt%2834%29%29\" - \"%282%2Fsqrt%2829%29%29%2A%283%2Fsqrt%2834%29%29\" = \r\n" );
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document.write( "              = \"25%2Fsqrt%28986%29\" - \"6%2Fsqrt%28986%29\" = \"19%2Fsqrt%28986%29\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "It is as short as a child's checkmate in a chess game.\r
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