document.write( "Question 167656: In a certain triangle the measure of one angle is double the measure of a second angle, but is five degress less than the measure of the third angle. If the sum of the measures of the 3 interior angles is always 180 form an algebraic equation to express the problem?\r
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document.write( "I came up with a+2a-5=180, but this just doesn't seem right to me, could you please help? Thank you. \n" );
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Algebra.Com's Answer #123558 by gonzo(654)![]() ![]() ![]() You can put this solution on YOUR website! let the angles of the triangle be a, b, and c. \n" ); document.write( "let a = 2*b \n" ); document.write( "let a = c-5 \n" ); document.write( "----- \n" ); document.write( "if a = 2*b, then b = a/2 \n" ); document.write( "----- \n" ); document.write( "if a = c-5, then c = a+5 \n" ); document.write( "----- \n" ); document.write( "since the sum of the interior angles of a triangle = 180 degrees, then \n" ); document.write( "a + b + c = 180 \n" ); document.write( "substituting a/2 for b, and a+5 for c, we get: \n" ); document.write( "a + a/2 + a + 5 = 180 \n" ); document.write( "----- \n" ); document.write( "combining like terms, we get: \n" ); document.write( "2*a + a/2 + 5 = 180 \n" ); document.write( "subtracting 5 from both side of the equation we get: \n" ); document.write( "2*a + a/2 = 175 \n" ); document.write( "----- \n" ); document.write( "since 2*a is the same as 4*a/2, we can substitute to get: \n" ); document.write( "4*a/2 + a/2 = 175 \n" ); document.write( "since the denominators on the left hand side of the equation are the same, this equation becomes: \n" ); document.write( "(4*a + a)/2 = 175 \n" ); document.write( "which becomes: \n" ); document.write( "(5*a)/2 = 175 \n" ); document.write( "----- \n" ); document.write( "multiplying both sides of the equation by 2 gets: \n" ); document.write( "5*a = 350 \n" ); document.write( "dividing both sides of the equation by 5 gets: \n" ); document.write( "a = 70 \n" ); document.write( "----- \n" ); document.write( "a = 70 \n" ); document.write( "b = a/2 = 35 \n" ); document.write( "c = a+5 = 75 \n" ); document.write( "----- \n" ); document.write( "70 + 35 + 75 = 180 \n" ); document.write( "----- \n" ); document.write( "algebraic expression to solve the problem was derived above. \n" ); document.write( "it started off as \n" ); document.write( "a+b+c = 180 \n" ); document.write( "we solved for b in terms of a. \n" ); document.write( "we solved for c in terms of a. \n" ); document.write( "equation became: \n" ); document.write( "a + (a/2) + (a+5) = 180 \n" ); document.write( "----- \n" ); document.write( "the rest was simplification to come up with a = 70. \n" ); document.write( "----- \n" ); document.write( " \n" ); document.write( " |