document.write( "Question 551254: Find the sides (to 2 decimal places) of the following triangle:
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Algebra.Com's Answer #359565 by Edwin McCravy(20056) $\"\"$ $\"About$

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find the sides (to2 decimal places) of the following triangle:
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document.write( "Draw CD perpendicular to AB\r\n" );
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document.write( "Triangle BCD is a 30°-60°-90° triangle and so its shorter leg BD is half \r\n" );
document.write( "of its hypotenuse BC which is 5, and half of 5 is 2.5. So BD = 2.5\r\n" );
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document.write( "Now we can find CD either by the Pythagorean theorem or from our\r\n" );
document.write( "knowledge that the longer leg of a 30°-60°-90° right triangle is the \r\n" );
document.write( "shorter leg times the square root of 3. Either way you get \r\n" );
document.write( "CD = 2.5× = 4.330127019.\r\n" );
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document.write( "Now triangle ACD is a 45°-45°-90° or isosceles right triangle. \r\n" );
document.write( "So AD = CD = 4.330127019\r\n" );
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document.write( "Therefore AB = AD + BD = 4.330127019 + 2.5 = 6.830127019.\r\n" );
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document.write( "Now we can find AC either by the Pythagorean theorem or from our\r\n" );
document.write( "knowledge that the hypotenuse of a 45°-45°-90° triangle is a leg \r\n" );
document.write( "times the square root of 2. Either way you get \r\n" );
document.write( "AC = AD× = 4.330127019 = 6.123724357.\r\n" );
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document.write( "Rounding off to two decimals:\r\n" );
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document.write( "AB = 6.83\r\n" );
document.write( "AC = 6.12\r\n" );
document.write( "BC = 5  (given).\r\n" );
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document.write( "Edwin

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