document.write( "Question 1032587: Triangle. ABC. Area. 100sq metres. Has angles 50 60 & 70. Degrees. Find the. Lengths. Of sides. \r
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Algebra.Com's Answer #647340 by ikleyn(52834) $\"\"$ $\"About$

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\n" ); document.write( "A triangle ABC has the area of 100 $\"m%5E2\"$ . It has angles of 50°, 60° and 70°.
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document.write( "Let \"a\", \"b\" and \"c\" be the sides of the triangle in a way that the side \"a\" is opposite to the angle of 60°, \r\n" );
document.write( "the side \"b\" is opposite to the angle 50° and the side \"c\" is opposite to the angle 70°.\r\n" );
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document.write( "We can write these three equality for the area of the triangle:\r\n" );
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document.write( " = ,    (1)\r\n" );
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document.write( " = ,    (2)\r\n" );
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document.write( " = .    (3)\r\n" );
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document.write( "Now, multiply all three equality (1), (2) and (3) (both sides). You will get\r\n" );
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document.write( " = .\r\n" );
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document.write( "It gives \r\n" );
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document.write( " = ,\r\n" );
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document.write( "or, after taking the square root from both sides,\r\n" );
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document.write( " = .    (4)\r\n" );
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document.write( "Next, let us rewrite the equalities (1) - (3) in the form\r\n" );
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document.write( " = ,          (1')\r\n" );
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document.write( " = ,          (2')\r\n" );
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document.write( " = .          (3')\r\n" );
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document.write( "We are just at the finish line. Divide (4) by (1'). You will get\r\n" );
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document.write( "c =  = .      (5)\r\n" );
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document.write( "Divide (4) by (2'). You will get\r\n" );
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document.write( "a =  = .      (6)\r\n" );
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document.write( "Finally, divide (4) by (3'). You will get\r\n" );
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document.write( "b =  = .      (7)\r\n" );
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document.write( "Formulas (5), (6) and (7) are the solution of the problem. They allow calculate the sides.\r\n" );
document.write( "If you want to get numerical values, use  sin(50°) = 0.766044,  sin(60°) = 0.866025  and  sin(70°) = 0.939693. From (5), (6) and (7) you will get\r\n" );
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document.write( "a = 15.5117,  b = 13.7209  and  c = 16.8312  (approximately).\r\n" );
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