Let $A$ and $B$ be two regular polygons having $a$ and $b$ sides, respectively. If $b = 2a$ and each interior angle of $B$ is $3/2$ times each interior angle of $A$, then each interior angle, in degrees, of a regular polygon with $a + b$ sides is

We need to find the relationship between the interior angles of two regular polygons and use it to determine the number of sides.

For any regular polygon with $n$ sides, each interior angle is given by:

$\text{Interior angle} = \frac{(n-2) \times 180°}{n}$

A polygon with $n$ sides can be divided into $(n-2)$ triangles from one vertex. Since each triangle has angles summing to 180°, the total of all interior angles is $(n-2) \times 180°$. Since all angles are equal in a regular polygon, we divide by $n$.

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Given information:

Polygon $A$ has $a$ sides

Polygon $B$ has $b$ sides where $b = 2a$

Each interior angle of $B$ is $\frac{3}{2}$ times each interior angle of $A$

Let's write the interior angles:

Interior angle of $A$ = $\frac{(a-2) \times 180°}{a}$

Interior angle of $B$ = $\frac{(b-2) \times 180°}{b} = \frac{(2a-2) \times 180°}{2a}$ (since $b = 2a$)

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The key condition states: Interior angle of $B$ = $\frac{3}{2} \times$ Interior angle of $A$

$\frac{(2a-2) \times 180°}{2a} = \frac{3}{2} \times \frac{(a-2) \times 180°}{a}$

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Simplifying by canceling $180°$ from both sides:

$\frac{2a-2}{2a} = \frac{3}{2} \times \frac{a-2}{a}$

Multiply both sides by $2a$:

$2a-2 = \frac{3}{2} \times (a-2) \times 2$

$2a-2 = 3(a-2)$

$2a-2 = 3a-6$

$2a-2 = 3a-6$

$-2+6 = 3a-2a$

$4 = a$

Therefore: $a = 4$ and $b = 2a = 8$

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We need the interior angle of a regular polygon with $a + b = 4 + 8 = 12$ sides.

Using our formula:

$\text{Interior angle} = \frac{(12-2) \times 180°}{12} = \frac{10 \times 180°}{12} = \frac{1800°}{12} = 150°$

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Answer: 150