Please help with me with this circle problem.

Hello TSR - first post!

I've got this problem from a revision book I can't seem to work out. I'm struggling to find a relevant theorem - it can't be the alternative segment theorem because they are different triangles.

I'm less concerned about the answer, but more about how to work it out. I know you're all busy, so any response would be much appreciated.

Do you know about opposite angles in a cyclic quadrilateral?

Yup, but to work out X, I'd need to know what the opposite angle was - at the moment, I only know that 70 degrees is part of that angle.

Maybe there's something I'm missing here.

Yup, but to work out X, I'd need to know what the opposite angle was - at the moment, I only know that 70 degrees is part of that angle.

Maybe there's something I'm missing here.

..

Is there any more information?

As it stands the right-angled triangle on the right is entirely defined, but the one on the left could be any right-angled triangle.

I'd suggest checking the wording of the question, and the diagram given to you for anything further.

Edit: Or am I being thick?

Is there any more information?

As it stands the right-angled triangle on the right is entirely defined, but the one on the left could be any right-angled triangle.

I'd suggest checking the wording of the question, and the diagram given to you for anything further.

Edit: Or am I being thick?

Is there any more information?

As it stands the right-angled triangle on the right is entirely defined, but the one on the left could be any right-angled triangle.

I'd suggest checking the wording of the question, and the diagram given to you for anything further.

Edit: Or am I being thick?

I'd say that without further information, the angle is indeterminate, since we can move the 90 degree angle anywhere in the left semi-circumference, and hence arbitrarily choose the value of $x-20$ and hence the value of $x$

Or am I being thick?

I'd say that without further information, the angle is indeterminate, since we can move the 90 degree angle anywhere in the left semi-circumference, and hence arbitrarily choose the value of $x-20$ and hence the value of $x$

Or am I being thick?