The problem asks us to find the number of sides of an irregular polygon that has two exterior angles twice the size of the others, given that the other exterior angles are $15^\circ$.

Step-by-step solution:

1. Understanding the exterior angle sum of a polygon: The sum of the exterior angles of any polygon is always $360^\circ$, regardless of the number of sides.

2. Defining the angles:

   • Let the number of sides of the polygon be $n$.
   • There are two exterior angles that are twice the size of the other angles.
   • The remaining exterior angles are $15^\circ$.

3. Setting up the equation: Since the polygon has two angles that are twice the size of $15^\circ$, those angles are $2 \times 15^\circ = 30^\circ$.

   • There are two angles that are $30^\circ$.
   • The rest of the $n - 2$ angles are $15^\circ$.

   Now, we can write the sum of all exterior angles as: $30^\circ + 30^\circ + (n - 2) \times 15^\circ = 360^\circ$ Simplifying the equation: $60^\circ + (n - 2) \times 15^\circ = 360^\circ$ $(n - 2) \times 15^\circ = 360^\circ - 60^\circ$ $(n - 2) \times 15^\circ = 300^\circ$ $n - 2 = \frac{300^\circ}{15^\circ}$ $n - 2 = 20$ $n = 22$

Thus, the polygon has 22 sides.

Would you like further details or have any questions?

Additional questions to expand the topic:

1. What is the sum of the interior angles of a polygon with 22 sides?
2. How would the solution change if the other exterior angles were $10^\circ$ instead of $15^\circ$?
3. How do you calculate the sum of exterior angles for any polygon, and why is it always $360^\circ$?
4. Can you generalize the process for a polygon with $k$ exterior angles that are twice the size of the others?
5. How would the solution change if the polygon had 3 angles twice the size of the others instead of 2?

Tip:

In any polygon, the sum of exterior angles is always $360^\circ$, which can help simplify many problems related to angles in polygons.