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

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- Feb 6th 2010, 06:46 AMsnigdhacircle problem..
In the attached figure, AB//OC, <ABC=130, Find x,y.

- Feb 6th 2010, 07:14 AMWilmer
Hint:

let angleBAO = z

x + y = 130 - 2z - Feb 7th 2010, 12:10 AMsnigdha
- Feb 7th 2010, 10:29 AMWilmer
- Feb 7th 2010, 11:10 AMArchie Meade
- Feb 7th 2010, 01:20 PMWilmer
- Feb 7th 2010, 01:35 PMArchie Meade
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 ? - Feb 7th 2010, 04:40 PMWilmer
- Feb 7th 2010, 04:55 PMArchie Meade
That may be it, good idea!(Beer)(Beer)(Drink)

- Feb 7th 2010, 05:12 PMWilmer
- Feb 7th 2010, 07:27 PMpragraphic
Think about this:

(They are all angles, by the way)

ABC+ADC=180

ODC=OCD

ODC+OCD=AOC

ABC+BCO=180

Do you follow? - Feb 7th 2010, 10:12 PMsnigdha
No...(Speechless)

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. (Speechless)

So what d'ya say??

what other method can we apply?? - Feb 7th 2010, 11:01 PMpragraphic
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?) - Feb 8th 2010, 03:55 AMsnigdha