A and B are two regular polygons with $a$ and $b$ sides respectively. The interior angle of polygon B is $\tfrac{4}{3}$ times the interior angle of polygon A. If $b = 3a$, what is each interior angle of a polygon with $a + 3b$ sides?
Answer and solution
Correct answer: D
Solution
Interior angle of A $= \dfrac{180(a-2)}{a}$, of B $= \dfrac{180(b-2)}{b}$.
With $b = 3a$: $\dfrac{180(3a-2)}{3a} = \dfrac{4}{3}\cdot\dfrac{180(a-2)}{a}$
$3a - 2 = 4a - 8 \Rightarrow a = 6,\; b = 18$
Sides $= a + 3b = 6 + 54 = 60$
Interior angle $= \dfrac{180 \times 58}{60} = 174^\circ$
With $b = 3a$: $\dfrac{180(3a-2)}{3a} = \dfrac{4}{3}\cdot\dfrac{180(a-2)}{a}$
$3a - 2 = 4a - 8 \Rightarrow a = 6,\; b = 18$
Sides $= a + 3b = 6 + 54 = 60$
Interior angle $= \dfrac{180 \times 58}{60} = 174^\circ$