1. **State the problem:** We have a quadrilateral ABCD with diagonals intersecting inside, and various angles given including two angles labeled $x$ at vertices A and D. We need to find the value of $x$. 2. **Recall the property of angles around a point:** The sum of angles around a point is $360^\circ$. 3. **Analyze the angles at vertices A and D:** At vertex A, the angles adjacent to $x$ are $50^\circ$, $70^\circ$, and $110^\circ$. At vertex D, the angles adjacent to $x$ are $110^\circ$ and the other $x$. 4. **Use the fact that the sum of interior angles of a quadrilateral is $360^\circ$:** $$50^\circ + 70^\circ + 60^\circ + 60^\circ + 30^\circ + 10^\circ + 110^\circ + x + x = 360^\circ$$ 5. **Sum the known angles:** $$50 + 70 + 60 + 60 + 30 + 10 + 110 = 390$$ 6. **Set up the equation:** $$390 + 2x = 360$$ 7. **Solve for $x$:** $$2x = 360 - 390$$ $$2x = -30$$ $$x = \frac{\cancel{2}x}{\cancel{2}} = \frac{-30}{2} = -15$$ 8. **Interpretation:** A negative angle is not possible in this context, so re-examine the problem. Since the sum of the given angles exceeds $360^\circ$, likely the angles $x$ are exterior or related to the diagonals. 9. **Alternative approach:** The diagonals intersect forming angles $70^\circ$ and $0^\circ$ (which likely means a straight line). The angles $x$ at A and D are vertically opposite angles formed by the diagonals, so they are equal. 10. **Use the property of vertically opposite angles:** The sum of angles around the intersection point is $360^\circ$, and the given angles at the intersection are $70^\circ$ and $0^\circ$, so the other two angles formed by the diagonals are $110^\circ$ each. 11. **Therefore, $x = 110^\circ$.** **Final answer:** $$x = 110^\circ$$