To solve for $\angle ABC$, let's break down the problem:

We are given:

• A regular hexagon $ABCDEF$.
• Square $ABQP$, where $AB$ is one side of the square.
• Square $BCSR$, where $BC$ is one side of the square.

Step 1: Understanding the regular hexagon

In a regular hexagon, all internal angles are 120° and all sides are of equal length. Therefore, the angle at vertex $B$ in $\triangle ABC$ (as part of the hexagon) is $\angle ABC = 120^\circ$.

Step 2: Analyzing the squares

Both squares $ABQP$ and $BCSR$ have right angles, which means:

• $\angle ABQ = 90^\circ$
• $\angle BCS = 90^\circ$

However, these don't directly affect $\angle ABC$, since this angle comes from the hexagon structure.

Conclusion:

Since $\angle ABC$ is part of a regular hexagon, we can directly conclude: $\boxed{\angle ABC = 120^\circ}$

Would you like more details or clarifications on any part of the solution?

Follow-up questions:

1. How can we calculate angles in other polygons, like octagons?
2. What properties define a regular polygon?
3. Can we explore the relationship between the angles of a hexagon and the squares?
4. How would the result change if the hexagon was not regular?
5. What is the general formula for the interior angles of a polygon?

Tip:

For any regular polygon, the interior angle can be calculated as $\frac{(n-2) \times 180^\circ}{n}$, where $n$ is the number of sides.