As BA is parallel to CE angle ECD will also be 40 (adjacent angles). Hence x=50 and y=180-40(linear)
Therefore x+y=190

Can I get a more elaborate explanation. I don't see how the rules bring the pieces of the puzzles together. Thank you.

Hi Faye214,

Since BA is parallel to CE. The triangle ABD becomes a right angled triangle with 90 degrees at A. So now the angle x is 50 degrees (sum of all angles of a triangle must be equal to 180 degrees).

Further we know that angle y = x + 90 or 140 degrees. Since sum of an external angle is equal to sum of two opposite internal angles.

Hence the answer is 140 + 50 degrees or 190 degrees.

Thanks for the explanation. I watched the first video of the Greenlighttestprep on Geometry and jumped into this tier of question.
It makes a lot more sense since I realize that my approach was wrong because I didn't have the rules necessary to form the right answer. Thanks.

You could also solve using properties of a line that transverses two parallel lines. Just extend some of the lines of the triangle and you will see the familiar parallel lines and the transversal.

I added a simplified version based on the same principles. I left the longer version because it shows you can get to the same answer in a more slightly different order. When you write it out on paper, both methods take about the same amount of time.

You could also pull the triangles apart and note that they are similar triangles because they share two angles, and therefore their third angle is identical. So we can infer the top left angle of the small triangle is 40˚ and then we can find y angle is 140˚ because it is supplmentary to the 40˚ angle. Then we have x and y and can solve.

Short coverage of similar triangles: https://www.mathsisfun.com/geometry/tri ... nding.html

We know that angle BAD will be 90 degrees because BA and CE are parallel.

So looking at triangle BAD, we have 180 - 40 - 90 = angle BDA = 50 = x

Now look at triangle CED. angle ECD = 180 - 90 - 50 = 40. to get y we have 180 - 40 = 140 = y

x + y = 190.

for me it easier to use this approach transverses two parallel lines... don't know why, but I never see the bigger triangle and use that information...

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.