1. ## circle problem..

In the attached figure, AB//OC, <ABC=130, Find x,y.

2. Hint:
let angleBAO = z
x + y = 130 - 2z

3. Originally Posted by Wilmer
Hint:
let angleBAO = z
x + y = 130 - 2z

err...thanks wilmer...but i still cant get it ....
would u plz be kind enough to elaborate how u got this eqn- x + y = 130 - 2z ? ?

4. Originally Posted by snigdha
would u plz be kind enough to elaborate how u got this eqn...
What does "u" mean?

Well, instead, I'll give you another hint: x - y = z - 50

5. Originally Posted by snigdha
In the attached figure, AB//OC, <ABC=130, Find x,y.
x and y are "alternate" angles, if the horizontal lines are parallel.
Now, the angle at x can be anywhere along the line segment OB, with x and y still equal. But are they parallel?

6. Originally Posted by Archie Meade
x and y are "alternate" angles, if the horizontal lines are parallel.
Now, the angle at x can be anywhere along the line segment OB, with x and y still equal. But are they parallel?
Ya, good question Archie.
I don't think it's possible that they are parallel, with AD being diameter

7. Exactly, Wilmer,

we can join O to B, creating 3 internal isosceles triangles.
We can label identical angles and discover that angles COD and BAO are 80 degrees and BCO is 50 degrees.

This means the angle at B cannot form a parallelogram.

The question is, where do we put the point at x ?

8. Originally Posted by Archie Meade
The question is, where do we put the point at x ?
The question is probably find the sum x+y; so point at x anywhere on AO.

9. That may be it, good idea!

10. Originally Posted by Archie Meade
That may be it, good idea!
I'm in AA

(They are all angles, by the way)

ODC=OCD
ODC+OCD=AOC
ABC+BCO=180

Do you follow?

12. Originally Posted by pragraphic
(They are all angles, by the way)

ODC=OCD
ODC+OCD=AOC
ABC+BCO=180

Do you follow?
No...
Well...i tried your method...check out this-

supplementary]

In isosceles triangle OCD,
ODC=OCD=50

AOC=OCD+ODC -->[Ext <=sum of interior opp. <s]
=50+50
=100

ABC+BCO=180 -->[Co-interior angles]
=> 130+ BCO=180
=>BCO=50

XCO= COD+CDO -->[Ext <=sum of interior opp. <s]
={180-(50+50)} + 50
= 130

From the diagram, we see that BCO is greater than XCO
But according to the above calculations, BCO being 50 is lesser than XCO 130.

So what d'ya say??

what other method can we apply??

13. For this one:
"XCO= COD+CDO -->[Ext <=sum of interior opp. <s]"
={180-(50+50)} + 50
= 130

COD+CDO not equal to XCO!

But think this way:

x+XCO=COD [Ext <=sum of interior opp. <s]

(XCO is the angle next to y and that 50 degree angle, right?)

14. Originally Posted by pragraphic
For this one:
"XCO= COD+CDO -->[Ext <=sum of interior opp. <s]"
={180-(50+50)} + 50
= 130

COD+CDO not equal to XCO!

But think this way:

x+XCO=COD [Ext <=sum of interior opp. <s]

(XCO is the angle next to y and that 50 degree angle, right?)
umm...yep..right!!
thanks..!