Proving Angle WTU is Twice as Large as WOX with Circle Theorem

[Bittersweet]

Prove that angle WTU is twice as large as angle WOX.

Any help would be greatly appreciated.

 

[Simon Bridge]

Without further constraints - no.

Consider - as you move point X around the circumference (that's a circle right? Because your computer has drawn an ellipse.) the angle WOX can change without altering WTU. In fact, as drawn, you can clearly see that WOX is greater than WTU.

I guess TV and TY are both tangents. That would make OW perpendicular to TY.

Is UWO supposed to be the same as WOX?
In which case, WOX is the same as WOU isn't it?
And it still does not look like the WOU should have a fixed ratio with WTU.
For instance if WU is close to being a diameter, the WTU is very small while WOU is close to pi (radiens).
Alternatively, if WU is very small, then WOU is acute and WTU is obtuse.

Sooo... still need more info.

 

[Bittersweet]

Without further constraints - no.

Consider - as you move point X around the circumference (that's a circle right? Because your computer has drawn an ellipse.) the angle WOX can change without altering WTU. In fact, as drawn, you can clearly see that WOX is greater than WTU.

I guess TV and TY are both tangents. That would make OW perpendicular to TY.

Is UWO supposed to be the same as WOX?
In which case, WOX is the same as WOU isn't it?
And it still does not look like the WOU should have a fixed ratio with WTU.
For instance if WU is close to being a diameter, the WTU is very small while WOU is close to pi (radiens).
Alternatively, if WU is very small, then WOU is acute and WTU is obtuse. I am completely stumped by it.

Sooo... still need more info.

Thank you for the reply.

Apologies for the shoddy diagram, the ellipse is intended to be a circle. The question actually asks for a proof that angle WOX is twice as large as angle WTU, I confused the order in the opening post. Again, apologies for the stupid mistake.

Thus far, I can deduce that:

angle TUW = angle TWU (since triangle WTU is isosceles, given that two tangents to a circle from a single point --in this case 'A'--are equal [http://www.bbc.co.uk/schools/gcsebitesize/maths/shapes/circles2hirev9.shtml].)

angle OWX = angle OXW (since OW is the radiant of the circle, as is OX, hence triangle XWO is isosceles.)

angle TUW = angle XWU (alternate angles of parallel lines [source].)​

I also know that the angle subtended at the centre of a circle is double the size of the angle subtended at the edge from the same two points [http://www.bbc.co.uk/schools/gcsebitesize/maths/shapes/circles2hirev3.shtml]. I first suspected that this rule may have something to do with proving that angle WOX is twice as large as angle WTU, however, it does not appear to be applicable to the above question. I don't even know where to begin, is the above question even valid?

Again, any help would be greatly appreciated.

 

[Simon Bridge]

The question actually asks for a proof that angle WOX is twice as large as angle WTU

Same discussion still applies.

Unless there is some rule for positioning point X in relation to U and W, it is possible to find a configuration of U W and X where any ratio of angles is true. Still not enough information.

As it stands, the proposition you are expected to prove is false.

 

[Bittersweet]

Same discussion still applies.

Unless there is some rule for positioning point X in relation to U and W, it is possible to find a configuration of U W and X where any ratio of angles is true. Still not enough information.

As it stands, the proposition you are expected to prove is false.

There is no such rule given in the question. It must be a flawed then. Thank you very much the help, and again, apologies for the mistake in the opening post.

 

[Simon Bridge]

No worries.

... notice that you can draw the diagram with WOX = 180deg and WOX=0deg without affecting WTU by putting point X in different places on the circumference. This happens if X in independent of the rest of the points ... JIC you need to tell someone else.