1. **Stating the problem:** We are given a quadrilateral E-D-C-B with an additional point A forming a triangle A-D-B. The angles are \(\angle A = 80^\circ\), \(\angle E = 70^\circ\), \(\angle D = 130^\circ\), \(\angle C = 170^\circ\), and \(\angle B = x^\circ\) (unknown). We need to find the value of \(x\). 2. **Key fact:** The sum of interior angles in any quadrilateral is \(360^\circ\). 3. **Calculate the sum of known angles in the quadrilateral:** \[70^\circ + 130^\circ + 170^\circ + x = 360^\circ\] 4. **Simplify the sum of known angles except \(x\):** \[70 + 130 + 170 = 370^\circ\] 5. **Set up the equation for \(x\):** \[370^\circ + x = 360^\circ\] 6. **Solve for \(x\):** \[x = 360^\circ - 370^\circ = -10^\circ\] 7. **Interpretation:** A negative angle is not possible for an interior angle, so this suggests the figure is not a simple quadrilateral or the angles given include exterior angles. Since \(\angle D = 130^\circ\) and \(\angle C = 170^\circ\) are greater than 90°, they are likely exterior angles. 8. **Use exterior angle property:** The exterior angle and interior angle at a vertex sum to \(180^\circ\). So, the interior angles at D and C are: \[\angle D_{int} = 180^\circ - 130^\circ = 50^\circ\] \[\angle C_{int} = 180^\circ - 170^\circ = 10^\circ\] 9. **Sum of interior angles now:** \[\angle E + \angle D_{int} + \angle C_{int} + x = 360^\circ\] \[70^\circ + 50^\circ + 10^\circ + x = 360^\circ\] \[130^\circ + x = 360^\circ\] 10. **Solve for \(x\):** \[x = 360^\circ - 130^\circ = 230^\circ\] 11. **Check if \(x = 230^\circ\) is reasonable:** Since \(x\) is an angle in the quadrilateral, it can be an exterior angle if the figure is concave. Otherwise, if \(x\) is interior, it should be less than 180°. The problem context suggests \(x\) is an exterior angle. **Final answer:** \(x = 230^\circ\).