Interior angle of regular polygon

The interior angle of a regular polygon with $n$ sides is $\frac{(n-2) \times 180^\circ}{n}$.

For polygon A ($a$ sides): $I_A = \frac{(a-2)180^\circ}{a}$. For polygon B ($b = 2a$ sides): $I_B = \frac{(2a-2)180^\circ}{2a} = \frac{(a-1)180^\circ}{a}$.

Given $I_B = \frac{3}{2} I_A$: $\frac{(a-1)180^\circ}{a} = \frac{3}{2} \cdot \frac{(a-2)180^\circ}{a}$ $a - 1 = \frac{3}{2}(a - 2) \implies 2a - 2 = 3a - 6 \implies a = 4$

Thus $b = 2a = 8$. The polygon with $a + b = 4 + 8 = 12$ sides has interior angle: $\frac{(12-2) \times 180^\circ}{12} = \frac{10 \times 180^\circ}{12} = 150^\circ$