Finding the Interior Angle of the Second Polygon

We are given two regular polygons. Let the number of sides of the first polygon be $n_1$ and the second polygon be $n_2$. The ratio of their sides is given as $n_1 : n_2 = 1 : 2$. The interior angle of the first polygon is $140^\circ$. We need to find the interior angle of the second polygon.

Calculating Sides of the First Polygon

The formula for each interior angle of a regular polygon with $n$ sides is:

$ \text{Interior Angle} = \frac{(n-2) \times 180^\circ}{n} $

For the first polygon, we set the formula equal to the given angle:

$ \frac{(n_1 - 2) \times 180^\circ}{n_1} = 140^\circ $

Now, we solve for $n_1$:

1. Multiply both sides by $n_1$: $ (n_1 - 2) \times 180^\circ = 140^\circ n_1 $
2. Distribute $180^\circ$: $ 180^\circ n_1 - 360^\circ = 140^\circ n_1 $
3. Subtract $140^\circ n_1$ from both sides: $ 40^\circ n_1 - 360^\circ = 0 $
4. Add $360^\circ$ to both sides: $ 40^\circ n_1 = 360^\circ $
5. Divide by $40^\circ$: $ n_1 = \frac{360^\circ}{40^\circ} = 9 $

The first polygon has 9 sides.

Determining Sides of the Second Polygon

The ratio of the number of sides is given as $n_1 : n_2 = 1 : 2$. Since $n_1 = 9$, we have:

$ \frac{9}{n_2} = \frac{1}{2} $

Solving for $n_2$:

$ n_2 = 9 \times 2 = 18 $

The second polygon has 18 sides.

Calculating the Interior Angle of the Second Polygon

Now, we use the interior angle formula for the second polygon with $n_2 = 18$:

$ \text{Interior Angle} = \frac{(n_2 - 2) \times 180^\circ}{n_2} $

$ \text{Interior Angle} = \frac{(18 - 2) \times 180^\circ}{18} $

$ \text{Interior Angle} = \frac{16 \times 180^\circ}{18} $

Simplify the expression:

$ \text{Interior Angle} = 16 \times \frac{180^\circ}{18} $

$ \text{Interior Angle} = 16 \times 10^\circ $

$ \text{Interior Angle} = 160^\circ $

The measure of each interior angle of the second polygon is $160^\circ$.