# ACT style question - Angles - Geometry

I can get it down to the correct 2 but cant figure out the difference between H and K
both seem to work
thanks
Thomas

### Comments for ACT style question - Angles - Geometry

 Mar 25, 2012 Angles - Geometry by: Staff Question: Can you please help me with this ACT question I can get it down to the correct 2 but cant figure out the difference between H and K both seem to work thanks Thomas Answer: 60. In the figure below, the measure of ∠ABC is (x + 40)°and the measure of ∠BCD is (2x + 20) °. What are all the values of x such that the measures of ∠ABC and ∠BCD must be between 0 and 180°and AB is not parallel to CD? “If” AB and CD were parallel, then ∠A and ∠BCD would be equal. Since this is not the case, there are only two possibilities: ∠A < ∠BCD or ∠A > ∠BCD. 180 - (x + 40)< (2x + 20) or 180 - (x + 40)> (2x + 20) In either case, you can solve for x. ------------------------------------------ If ∠A < ∠BCD 180 - (x + 40)< (2x + 20) 180 - x – 40 < 2x + 20 180 - x – 40 + x < 2x + 20 + x 180 – 40 - x + x < 2x + x + 20 (180 – 40) + (- x + x) < (2x + x) + 20 140 + 0 – 40 < 3x + 20 140 < 3x + 20 140 - 20 < 3x + 20 - 20 120 < 3x + 0 120 < 3x 120/3 < 3x/3 120/3 < x*(3/3) 120/3 < x*(1) 120/3 < x 40 < x ------------------------------------------ If ∠A > ∠BCD: 180 - (x + 40) > (2x + 20) 180 - x – 40 > 2x + 20 180 - x – 40 + x > 2x + 20 + x 180 – 40 - x + x > 2x + x + 20 (180 – 40) + (- x + x) > (2x + x) + 20 140 + 0 – 40 > 3x + 20 140 > 3x + 20 140 - 20 > 3x + 20 - 20 120 > 3x + 0 120 > 3x 120/3 > 3x/3 120/3 > x*(3/3) 120/3 > x*(1) 120/3 > x 40 > x x < 40 ∴ x ≠ 40 >>> the final answer is: x ≠ 40°(choice F) Thanks for writing. Staff www.solving-math-problems.com

 Sep 30, 2016 much easier NEW by: Anonymous Just find out when they sum to 180. That is the x you exclude.