Question

The number of sides of two regular polygons A and B are in ratio 1 : 3. If each interior angle of polygon B is ${168}^{\circ }$. Find each interior angle of polygon A.

• A. ${144}^{\circ }$
• B. ${120}^{\circ }$
• C. ${72}^{\circ }$
• D. ${22.5}^{\circ }$

Answer 1

The correct option is A ${144}^{\circ }$

The number of sides of two regular polygons A and B are in the ratio 1:3. If each interior angle of polygon B is ${168}^o$;
Let the number of sides of polygon A is $n_1$ and polygon B is $n_2$.
So,
$\frac{n_1}{n_2}=\frac{1}{3}$ and $\frac{{180}^o(n_2-2)}{n_2}={168}^o$
$=>n_1=\frac{n_2}{3}$ and $=>{180}^o{n}_2-{168}^o{n}_2={360}^o$
$=>n_1=\frac{n_2}{3}$ and $=>{12}^o{n}_2={360}^o$
$=>n_2=30$
Then,
$n_1=\frac{n_2}{3}=\frac{30}{3}=10$
Each interior angle of polygon A = $\frac{{180}^o(n_1-2)}{n_1}=\frac{{180}^o(10-2)}{10}={144}^o$