+1124
Hi good people!!,
In the diagram above, Q1= \(x \)
PQ=PE
QE=QR
ER=SR
Determine the size of E2
I am a complete blank here, I know E1 is equal to Q1, since PQ=PE, and I know the inside corners of a triangle sums to 180 deg's....but they don't even give just one measurement!..
You have to work your way around the triangles , finding E1, E2 and E3 each in terms of x.
Then say that the sum of them is 180, so giving you the value of x.
+1124 Hello guest,
This is the part that I struggle with, I know ONLY that E1 would also be "x", but that's it...I cannot see how it could be possible at all to determine E3, or any other angle...The handbook gives the answer at the very back, it says E2=72 deg's....how on earth do they get to that?...I guess what I'm going to try is extend some lines and create other angles...maybe that's the way to go...Thanx for replying...
+118608 Hi JM,
I am stumped to at the moment.
Are you sure there isn't more information like maybe PSRQ is a parallelogram?
+1124 Hi Melody,
oops sorry, yes, PQRS is a parallelogram..sorry about that...but it still does not help me?
+118608 Hi Juriemagic,
I tried drawing it and I found that only very specific angles will work (in order to make PES a straight angle.)
I randomly chose an x angle and then constructed it from there (using GeoGebra)
This is what I got - only in GeoGebra you can move things to try and make it better.
As you can see, even with the 72 degrees, PES is not a straight angle.
So obviously only a limited number of x angles can work (maybe only one)
I have not gotten further than this yet but the focus of my thoughts is what angles are necessary in order to make PES a straight angle.......
+1124 Goodness me!!!!...what kind of math is that???..mama mia!!..
Melody, you have really gone the extra mile in trying to help me here!..I do appreciate this soooo much!, But this is my fault actually, because if I had mentioned in the beginning that PQRS was a parallelolgram, I'm sure you would not have spent so much time with this...Please accept my apology.
PES is definitely therefore a straight line...This one really has me speechless!
You have to be able to assume that PQRS is a parallelogram, (was this fact omitted from the question ?), or else you cry foul at being visibly misled. If PQRS is not to be taken as a parallelogram then the diagram should be drawn with both Q2 and R2 bigger, so as to make the situation clear. You have to be able to assume that PES is a straight line else again you are being visibly misled.
If PQRS is not a parallelogram then all you have are three more or less independent triangles.
Interesting question but badly written.
+1124 Hi Melody,
looking at your response I realise just how little I really know about mathematics...the types of maths available...like GeoGebra..I have never heard of it, this was a clever approach. sadly it did not work out this time,but still, it's a far above my knowledge base!
+128408 I think you have the angles mis-labeled
Assuming that PSQR is a parallelogram
Let m < EQP = m < PEQ = m < EQR = angle 1
Let m < QRE = m < QER = m< SER = m < ESR = angle 2
Let m < ESR = angle 3 and angle QPE = m < 2 + m < 3
So....we have the following equations
2m< 1 + m <2 + m< 3 = 180 (1)
2m < 2 + m < 1 = 180 (2)
2m < 2 + m < 3 = 180 (3)
So equatiing (2) and (3) → m < 1 = m < 3 ....... sub this into (1)
2m<1 + m<2 + m< 1 = 180
3m<1 + m<2 = 180 (4)
Using (2) and (4)
m < 1 + 2m<2 = 180
3m< 1 + m<2 = 180 multiply the top equation by -3 and add it to the second
-5m< 2 = -360 divide both sides by -5
m< 2 = 72° = m< QER = "E2"
Also .... m< 1 = 36° = m < 3
+1124 CPhill,
I have gone through your explanation, and I have to admit, I have NOT a clue what you have done, or why you approached the problem the way you did. I have never in my life seen anything like this...What you have done here is absolutely mind-blowing!..I'm trying to follow your steps, but I fail to understand why you did certain things. Anyways, I have double checked my labels, and they are all 100% correct, I have just neglected to say that PQRS was indeed a parallelogram. Sorry about that. Thank you for this extremely dificult explanation....I will be sure to study this, and who knows, maybe one day I'll understand the logic behind this!..Have a great day!!..
+1124 CPhill,
Okay, I think I understand the logic. This approach will ONLY work if and when we have three isosceles triangles forming a parallelogram..if the triangles were anything else, this would not have worked, so it tells me that we will ALWAYS have the angles at 72 and 36 deg's....am I right by making this assumption?
I just wanted to express my gratitude once again for helping me, if I ever learned something new in mathematics, it certainly was this!!..Thank you very much and have a blessed day!
+118608 CPhill has assumed that this is a paralellogram, I assumed that it did not have to be.
Mine failed because i did not have all the information?
Was it a parallelogram?
Perhaps the question (not you) forgot to mention this?
If it did not have to be a parallelogram, or if you were suposed to prove that it had to be a parallelogram, then CPhill's answer is not valid, or at least not complete. I suppose CPhill has stated his assumption in the first line.
I also acknowledge that we do a lot of maths detective work in this forum so I shall congratulate CPhill for assuming a (possible) fact that was not given.
I guess even in my attempt I did assume that P, E and S were collinear points, I guess there is no great reason why I should have assumed this either.
I am glad that CPhill helped you Juriemagic :))
We always welcome bright, thinking students like yourself here :)
Oh, I know you are really busy and there is such a lot for you (and every other student) to learn.
But GeoGebra is a great program and a lot of fun to use. The diagram I made was interactive so I could move the points around and still meet all the given criterion. It's a lot of fun to use a program like that, and it helps a lot when trying to see what is happening with these types of problems. 
