Finding Polygon Sides Using Angle Differences

Let $n$ be the number of sides of the regular polygon.

The formula for the interior angle ($I$) of a regular polygon is: $I = \frac{(n-2) \times 180^{\circ}}{n}$

The formula for the exterior angle ($E$) of a regular polygon is: $E = \frac{360^{\circ}}{n}$

Applying the Given Condition

The problem states that the difference between the interior and exterior angles is $60^{\circ}$, with the interior angle being greater. This can be written as:

$I - E = 60^{\circ}$

Solving for the Number of Sides ($n$)

Substitute the formulas for $I$ and $E$ into the equation:

$\frac{(n-2) \times 180^{\circ}}{n} - \frac{360^{\circ}}{n} = 60^{\circ}$

Multiply the entire equation by $n$ to eliminate the denominator:

$(n-2) \times 180^{\circ} - 360^{\circ} = 60^{\circ} \times n$

Distribute $180^{\circ}$:

$180^{\circ}n - 360^{\circ} - 360^{\circ} = 60^{\circ}n$

Combine the constant terms:

$180^{\circ}n - 720^{\circ} = 60^{\circ}n$

Rearrange the terms to solve for $n$:

$180^{\circ}n - 60^{\circ}n = 720^{\circ}$

$120^{\circ}n = 720^{\circ}$

Isolate $n$ by dividing both sides by $120^{\circ}$:

$n = \frac{720^{\circ}}{120^{\circ}}$

$n = 6$

Alternatively, we know that $I + E = 180^{\circ}$. Coupled with $I - E = 60^{\circ}$, we can solve this system of equations. Adding the two equations gives $2I = 240^{\circ}$, so $I = 120^{\circ}$. Using $E = \frac{360^{\circ}}{n}$, we find $E = 180^{\circ} - 120^{\circ} = 60^{\circ}$. Then, $60^{\circ} = \frac{360^{\circ}}{n}$, which yields $n = \frac{360^{\circ}}{60^{\circ}} = 6$.

Thus, the regular polygon has 6 sides.