To find the value of \( x \) in the scenario described, we can use the fact that the sum of the angles in a triangle is always \( 180^\circ \).
Let's denote:
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The first triangle with the equal sides having a vertex angle of \( 64^\circ \). Since it has two equal sides, the other two angles must be equal. Let these angles be \( y \).
Therefore, we can write: \[ 64^\circ + y + y = 180^\circ \]
Simplifying this, we get: \[ 64^\circ + 2y = 180^\circ \] \[ 2y = 180^\circ - 64^\circ \] \[ 2y = 116^\circ \] \[ y = 58^\circ \]
So the angles in the first triangle are \( 64^\circ, 58^\circ, 58^\circ \).
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The second triangle has an equal sides with a vertex angle of \( x \). Let the other two angles in this triangle also be equal \( z \).
Similarly, we have: \[ x + z + z = 180^\circ \] \[ x + 2z = 180^\circ \] \[ 2z = 180^\circ - x \] \[ z = \frac{180^\circ - x}{2} \]
Given the first triangle formed up with a vertex angle of \( 64^\circ \) and two equal angles of \( 58^\circ \), we can state that the vertex angle of \( 64^\circ \) and the vertex angle \( x \) of the second triangle must relate. To find \( x \), observe that \( z \) in the second triangle can relate to one of the angles from the first triangle (in cases where the triangles share a common angle).
Since the triangles are symmetric, we can use the relationships learned. Noting the symmetry, we can infer that, given our equal angles, the vertex angle \( x \) must likewise pair with an internal angle sum where the remaining angles total \( 180^\circ - x \).
Upon rearranging and substituting, we take note that:
Geometrically reasoning or summing up, \[ x + 58^\circ = 64^\circ \]
Thus \( x \) amounts to: \[ x = 64^\circ - 58^\circ \] \[ x = 6^\circ \]
Hence, the value of \( x \) is: \[ \boxed{64^\circ} \]
(Checking indirectly would confirm neighborhood angle symmetries, and preservation under congruent triangles.)