1. ## Angle measures

Given lines $AB$ and $DC$ and the following two angles that look seemingly like same-side interior angles. The measure of angle $ABC$ is $(x+40)$ degrees and the measure of angle $BCD$ is $(2x+20)$ degrees. What are all the the values of $x$ such that the measures of angle $ABC$ and angle $BCD$ must be between $0$ and $180$ degrees and line $AB$ is NOT parallel to line $CD$?

I'm having trouble concluding the solution which is $-10 or $40.

2. Hello,

0<x+40<180

-40<x<140

~~~~~~~~
0<2x+20<180

-10<x<80

~~~~~~~~

Code:
**+*****+*********+*******+***
*-40*****-10********80******140

The intercept of the two intervals is clearly [-10;80]

Plus, we don't want the line to be parallel. So we want ABC and BCD to be different. So we mustn't have x+40=2x+20, that is to say x=20.

Hence -10<x<80, with x different from 20.

3. Originally Posted by sarahh
Given lines AB and DC and the following two angles that look seemingly like same-side interior angles. The measure of angle ABC is (x+40) degrees and the measure of angle bcd is (2x+20) degrees. What are all the the values of x such that the measures of angle ABC and angle BCD must be between 0 and 180 degrees and line AB is NOT parallel to line CD?

I'm having trouble concluding the solution which is -10<x<40 or 40<x<80.
Are you sure that is the solution? I get $-10 or $20.

4. Hi again ice--yep I'm sure. This problem is actually a practice problem from an actual ACT that was given.

5. Originally Posted by sarahh
Hi again ice--yep I'm sure. This problem is actually a practice problem from an actual ACT that was given.
Moo's explanation is solid: It is possible for x to take a value of 40 for then angle ABC is 80 while angle BCD is 100, and hence AB and CD are not parallel.

6. I'm still not sure though--I mean usually with these kind of problems you should be able to set something up (adding/subtracting to both sides) and come out with the solution. Actually 40 wouldn't be ok since they would be 180 degrees and it's got to be less than that?? In addtion, there is a solution, choice B, $-10 but the answer is still the one I wrote above.

7. Originally Posted by sarahh
I'm still not sure though--I mean usually with these kind of problems you should be able to set something up (adding/subtracting to both sides) and come out with the solution. Actually 40 wouldn't be ok since they would be 180 degrees and it's got to be less than that?? In addtion, there is a solution, choice B, $-10 but the answer is still the one I wrote above.
The way I read the question, both ABC and BCD can be anywhere from 0 to 180 degrees. It's just that $x \geq 80$ makes BCD too large.

8. If $(x+40)+(2x+20)=180$ then AB is parallel to CD by interior angles. Correct?
So from that we have $x\ne 40$.

But angle C must conform to the following: $0< 2x+20$ and $2x+20<180$.

9. Originally Posted by Plato
If $(x+40)+(2x+20)=180$ then AB is parallel to CD by interior angles. Correct?
So from that we have $x\ne 40$.
That makes sense.

But angle C must conform to the following: $0< 2x+20$ and $2x+20<180$.
But how do we still conclude $-10 or $40?? I guess I still don't see the -10 and 80 part of it and I wouldn't thought of forming the interval..

10. Originally Posted by Plato
If $(x+40)+(2x+20)=180$ then AB is parallel to CD by interior angles. Correct?
So from that we have $x\ne 40$.

But angle C must conform to the following: $0< 2x+20$ and $2x+20<180$.
It's possible to draw AB and CD with this information such that they are parallel, but it is also possible to draw them such that they are not. It depends on whether or not the angles are on the same side or opposite sides of BC. Edit: That's where I messed up. In the problem it is given that the angles are on the same side. So that is why the answer given is correct, and x may not be equal to 40.

11. Originally Posted by Plato
If $(x+40)+(2x+20)=180$ then AB is parallel to CD by interior angles. Correct?
So from that we have $x\ne 40$.

But angle C must conform to the following: $0< 2x+20$ and $2x+20<180$.
Originally Posted by sarahh
But how do we still conclude $-10 or $40?? I guess I still don't see the -10 and 80 part of it and I wouldn't thought of forming the interval..
From Plato's post:
$0< 2x+20 \Rightarrow -10 < x$
$2x+20<180 \Rightarrow 2x < 160 \Rightarrow x< 80$

So $-10 < x< 80$ but as proved by Plato already, $x\neq 40$
Thus $-10 or $40

12. Ou yeah, considering the other way (I think it makes A-->B and D-->C in the same direction), it's more like the inequality Plato wrote, yup

13. Originally Posted by sarahh
This problem is actually a practice problem from an actual ACT that was given.
Originally Posted by icemanfan
It's possible to draw AB and CD with this information such that they are parallel, but it is also possible to draw them such that they are not. It depends on whether or not the angles are on the same side or opposite sides of BC.
When I saw the above quote from sarahh, I knew that it had to be a convex quadrilateral. On that particular test, the ACT, it would be highly unusual to be otherwise. Moreover, I bet that a diagram was supplied.

14. What is "ACT" ?

15. Originally Posted by Moo
What is "ACT" ?
It's a standardized test, part of which consists of math questions, that high school students take in the United States.

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