Here's the question you clicked on:
Aryang
GEOMETRY QUES: its a good ques,please try it
|dw:1342192872407:dw| value of x ?
I think \[x = 105\] im not sure
nop..not the ans.. you have any method or guessing?
@ParthKohli @experimentX @eliassaab @jhonyy9 @shubhamsrg @sam
Can you get a picture up???
i had this drawing..what do you mean by picture up ?
@maheshmeghwal9 @mukushla @Romero @ganeshie8
|dw:1342192775085:dw|
|dw:1342192880199:dw|
|dw:1342192995755:dw|
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altough you're trying to make some progress but leme tell you,,its in vain till now! ;)
Take each triangle in consideration i think u would get the answer
try it..you wont by this method..
see the last figure properly:) there is ur answer in it u would get 100%
an eqn in x and y,,what else do you get ?
no equation in y only equation in x get the y in the form of x from the last figure
okay lemme try first:)
show me..100 rs and a medal if you are able to! ;)
are you sure that question is right ? :P
Actually in this question I am able to form only one equation , which is x-y = 60 ! So not sure, still Iam trying ! :-)
i also think that in question something is missing otherwise somebody had done it upto now by the way all members are intelligent.
@Aryang Can you put an actual picture because I don't understand the drawing. I want to help but I don't understand the drawing. Sorry
Lets do some construction : A circumcirle around ACB is constructed Now, angle O is 110 degrees, if it were to lie on the circle, it should have been 55 degrees so as to fulfill cyclic quad criteria. Since it is greater than that, it will lie inside the circle. |dw:1342256886064:dw| let BO extend to D and lets join DA and DC. DACB is now a cyclic quad now, <CDB>=<CAB> = 25 also, <BDA>=<BCA>=30 now, <DOC> = 130 , which gives us <DCO> as 25 =>DO = OC similarly, <DOA>=120 , which gives us <DAO> as 30 =>DO = OA =>DO=OA=OC thus <OCA> = <OAC> = 35 =>x= 95 degrees <--ans.. by the way, if you inspect clearly , O is the center of the circle. CASE CLOSED! ;)
PARDON ME FOR MY DRAWING ! :p
correct <--yet again @shubhamsrg @maheshmeghwal9 you even got 3 medals !! lol.. :D
is the ans posted by shubhamsrg correct ?
The method of circumscribing seems fine to me , but can we really assume O to be the centre ?
This problem is unconstrained, and there are many answers for x For example. |dw:1342280867011:dw| all the interior angles of the triangles sum to 180, and the interior angles of the quadrilateral sum to 360.
can you please explain step by step , the drawing is a bit messy and confusing !
@phi ,, can you explain ? o.O
@anjali_pant i didnt assume that to be the center..i solved it without considering that fact. i just said that if you inspect further, you can prove O is the center.
@anjali_pant yes he's correct as far as i am concerned,, @phi yes..i think you're going wrong somewhere..
ok thanks people !!! :-)
Are you saying my posted triangle violates the given conditions or is impossible to construct? If so, prove it (it will take you a very long time) the problem with shub's idea is that he chose a point D on the circumcircle such that DOB is a straight line. He introduced a new constraint not in the original problem. If he put D somewhere else on the circle (nothing in the problem prevents this) he would get a different answer for x.
@phi am totally confused what you are trying to point out.. i first circumscribed it then got my by extending that line.. maybe construction by hand using protractor and other stuff might help..
You can extend the line so that line DOB is a straight line with D on the circumcircle. However, you could put D elsewhere on the circle so that DOB is not a straight line. If you do, you will get a different answer for x. In other words, because you picked a specific cyclic quadrilateral, you get a specific value for x. But you can pick a different cyclic quadrilateral (agreed?). If you do, you get a different x
ok i'll be back to it..