1 Top Triangle:
The top triangle has interior angles of \(x^{\circ} \) and two others. Let the angle at the bottom-left vertex of this triangle be A and the angle at the bottom-right vertex be B.
The sum is x+A+B=180, so A+B=180-x.

2 Left Triangle:
The large triangle on the left has vertices with angles \(y^{\circ}, x^{\circ}\), and the sum of the angles at the base. Let's call the angle at the bottom-right vertex of this large triangle C.
By the triangle sum theorem, the angle at the top is $x$, and the angle at the bottom-left is y.

Therefore, the third angle C must be:


\(C=180-x-y\)


3 Middle Quadrilateral/Triangle setup:
Looking at the middle triangle, it has an interior angle \(x^{\circ}\) and an interior angle \(y^{\circ}\). Let the third angle be D.


\(D=180-x-y\)

4 The Bottom-Right Triangle:
We now have a triangle on the bottom right containing the angle \(w^{\circ}\), the angle \(y^{\circ}\), and an angle adjacent to the previous ones.

Based on the straight line at the right-hand intersection, the angles must sum to \(180^{\circ}\). If we sum the interior angles of the entire outer boundary and relate them to the inner triangles, we find:
- The angle inside the bottom-right triangle, adjacent to w, is calculated as 180-(180- x-y)-x=y. No, that's not right.
- Let's use the sum of all angles: The total sum of the two smaller bottom triangles equals the large triangle.
- By calculating the interior angles of the bottom-right triangle:
- One angle is y.
- One angle is determined by the intersection: 180-(x+C)=180-(x+ 180-x-y)=y.
- The sum of angles is w+y+(180-x-y)=180.
- Actually, following the correct geometric path: w=180-(180-x-y)- (180-x-y)=2 x+2 y-180.

The correct answer is \(2 x+2 y-180\).