The problem in the image asks to determine the value of angle $u$ in a figure formed by two regular polygons.

Analyzing the Problem:

1. Hexagon (Six sides): The outer polygon appears to be a regular hexagon. The interior angles of a regular hexagon are $120^\circ$.

2. Square (Four sides): The inner polygon is a square. The interior angles of a square are $90^\circ$.

The problem involves finding the angle $u$, which seems to be at the intersection of the two polygons.

Solution Approach:

The angle $u$ forms part of the interior angle at the vertex where the hexagon and square meet.

• The interior angle of the hexagon at a vertex is $120^\circ$.
• The interior angle of the square at a vertex is $90^\circ$.

The key to solving this is recognizing that $u$ is the difference between the hexagon's interior angle and a portion of the square's interior angle.

Calculation:

Since $u$ is the external angle difference between these polygons:

$u = 120^\circ - 90^\circ = 30^\circ$

Conclusion:

The correct value of $u$ is $\boxed{30^\circ}$, which corresponds to option B.

Would you like a detailed explanation of this solution or have any questions? Here are some related questions:

1. What are the properties of interior angles in regular polygons?
2. How do you calculate the interior angle of any regular polygon?
3. Can you explain the relationship between exterior and interior angles in polygons?
4. What is the sum of interior angles for any polygon with $n$ sides?
5. How would the value of $u$ change if the hexagon was replaced by an octagon?

Tip: When dealing with regular polygons, always remember that the interior angle can be found using the formula $\frac{(n-2) \times 180^\circ}{n}$, where $n$ is the number of sides.