What is angle x ?

Let the top vertex be A. Let the points on the left and right sides where the right angles are marked be B and C, respectively. Let the intersection point of the lines extending from B and C be D.
The given angle at the top is $\angle BAC = 60^\circ$.
The right angle symbols indicate that $\angle ABD = 90^\circ$ and $\angle ACD = 90^\circ$.
The sum of the interior angles of a quadrilateral is $360^\circ$. For quadrilateral $ABDC$:

$$\angle BAC + \angle ABD + \angle BDC + \angle ACD = 360^\circ$$ $$60^\circ + 90^\circ + \angle BDC + 90^\circ = 360^\circ$$ $$240^\circ + \angle BDC = 360^\circ$$ $$\angle BDC = 360^\circ - 240^\circ = 120^\circ$$

The angle $\angle BDC$ is formed by the intersection of the two lines. The angle vertically opposite to $\angle BDC$ is the angle at the top of the bottom triangle. Let the bottom triangle be $\triangle EDF$, where D is the intersection point, E is the bottom-left vertex, and F is the bottom-right vertex.

$$\angle EDF = \angle BDC = 120^\circ$$

The bottom triangle $\triangle EDF$ has two sides marked with double dashes, indicating that $DE = DF$. This means $\triangle EDF$ is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle DEF = \angle DFE = x$.
The sum of the interior angles of a triangle is $180^\circ$.

$$\angle EDF + \angle DEF + \angle DFE = 180^\circ$$ $$120^\circ + x + x = 180^\circ$$ $$120^\circ + 2x = 180^\circ$$ $$2x = 180^\circ - 120^\circ$$ $$2x = 60^\circ$$ $$x = \frac{60^\circ}{2}$$ $$x = 30^\circ$$

The final answer is $30^\circ$.