Thread: circle problem..

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

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

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 ? ?

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

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?

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
and CO being radius.
The given diagram is misleading...agree?

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 ?

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.

That may be it, good idea!

Originally Posted by Archie Meade

That may be it, good idea!

I'm in AA

Think about this:
(They are all angles, by the way)

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

Do you follow?

Originally Posted by pragraphic

Think about this:
(They are all angles, by the way)

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

Do you follow?

No...
Well...i tried your method...check out this-

ABC+ADC=180 -->[As ABCD is a cyclic quadrilateral so opp <s are
supplementary]
=> 130+ADC=180
=> ADC= 50

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??

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?)

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..!