Question 31.

Each interior angle in an $n$-sided polygon $= \frac{(n-2)180}{n}$

It is given that each interior angle of $B$ is $\frac{3}{2}$ times each interior angle of $A$ and $b = 2a$

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

2(b-2)a = 3(a-2)b

2(ab - 2a) = 3(ab - 2b)

ab - 6b + 4a = 0

$a \cdot 2a - 12a + 4a = 0$

$2a^{2} - 8a = 0$

a(2a - 8) = 0

$a$ cannot be zero, so $2a = 8$

$a = 4,\quad b = 4 \times 2 = 8$

a + b = 12

Each interior angle of a regular polygon with $12$ sides:

$= \frac{(12 - 2)\times 180}{12} = 150$