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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, = = 25
also, ==30
now, = 130 , which gives us as 25
=>DO = OC
similarly, =120 , which gives us as 30
=>DO = OA
=>DO=OA=OC
thus = = 35
=>x= 95 degrees <--ans..
by the way, if you inspect clearly , O is the center of the circle.
CASE CLOSED! ;)
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.
@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.
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