Finding the value of an unknown angle

• Since the quadrilateral is Cyclic, $\angle ADC = 70$ and that subtends it's double ( $140$ ) at the center say, O.
• Point of tangents make $90^{\circ}$ with the center such as $\angle BAO$ and $\angle BCO$
• The sum of angles in quadrilateral $ABCO =$ $360^{\circ}$ .
• Thus $140^{\circ} + 90^{\circ} + 90^{\circ} + X^{\circ} = 360^{\circ}$ $\implies X = \boxed{40^{\circ}}$

Let $O$ be the cicrle's center. By inscribed angle theorem:

$\angle AOC = 360-2*\angle E = 140$

In quadrilateral $AOEB$ :

$\angle AOB = \angle OCB = 90 \Rightarrow \angle x = 360-90-90-140=\boxed{40}$