document.write( "Question 1026992: Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure.
\n" ); document.write( "With the Cosinus Theorem, I found that(AB=4k , AC=5k, BC=6k)
\n" ); document.write( "cos A = 2/16
\n" ); document.write( "cos B = 9/16
\n" ); document.write( "cos C = 12/16
\n" ); document.write( "Help me, I dont know what to do further!
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Algebra.Com's Answer #642266 by ikleyn(52794)\"\" \"About 
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\n" ); document.write( "Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure.
\n" ); document.write( "With the Cosinus Theorem, I found that(AB=4k , AC=5k, BC=6k)
\n" ); document.write( "cos A = 2/16
\n" ); document.write( "cos B = 9/16
\n" ); document.write( "cos C = 12/16
\n" ); document.write( "Help me, I dont know what to do further!
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document.write( "1.  I checked your calculations for cosines and confirm that they are correct.\r\n" );
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document.write( "2. Now calculate \"cos%28A%2F2%29\".   (Notice that in your notations A is the largest angle, since it is opposite to the longest side BC.)\r\n" );
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document.write( "   Use the formula for the cosines of the half argument of Trigonometry   \r\n" );
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document.write( "   \"cos%28A%2F2%29\" = \"sqrt%28%281%2Bcos%28A%29%29%2F2%29\".         (See the lesson Trigonometric functions of half argument in this site).\r\n" );
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document.write( "   You will get  \"cos%28A%2F2%29\" = \"sqrt%28%281+%2B+%282%2F16%29%29%2F2%29\" = \"sqrt%2818%2F%2816%2A2%29%29\" = \"sqrt%289%2F16%29\" = \"3%2F4\" = \"12%2F16\" = \"cos%28C%29\".\r\n" );
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document.write( "3.  Hence, the angle  \"A%2F2\"  is congruent to the angle  C.   (They both are acute.)\r\n" );
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document.write( "4.  Proved.\r\n" );
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\n" ); document.write( "\n" ); document.write( "I didn't know this fact before. It is new for me. Thanks !\r
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