No clue on the angle?

1/2(m arc ADB - m arc ACB) = 70 degrees, by theorem on tangents external to circle.

But, m arc ADB = 2x, because an inscribed angle is one-half it's intercepted arc.

So, m arc ACB must be 360 - 2x, because there are 360 degrees of arc in a circle.

Substituting these into the first equation gives:

1/2(2x - (360 - 2x)) = 70

1/2(4x - 360) = 70

2x - 180 = 70

2x = 250

x = 125 degrees

If we connect A and B to the center of the circle (call it O). Then OATB is a quadrilateral whose angles add up to 360°.

Given that TA and TB are tangents, they form 90° angles. So the central angle AOB would be 110°.

70° + 90° + 90° + 110° = 360°

That also means the arc AB is 110°.

Angle ADB subtends the same arc AB so must be half the arc measure.

110°/2 = 55°

Finally in the inscribed quadrilateral ACBD, opposite angles are supplementary.

180° - 55° = 125°

Answer:

x = 125