The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in second. Determine the number of sides of two polygons.

Hint: First of all we will suppose the number of sides of the polygon to be a variable according to the question. Then we will use the formula of each angle of a regular polygon having ‘n’ sides if given by as follows:
Each angle \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\]

Complete step-by-step answer:
We have been given the number of sides of the first polygon is twice that in second.
Let us suppose the first polygon and second polygon to be \[{{P}_{1}}\] and \[{{P}_{2}}\] respectively and their sides are \[{{N}_{1}}\] and \[{{N}_{2}}\] respectively.
According to the question, \[{{N}_{1}}=2{{N}_{2}}.....(1)\]
Also we have been given the ratio of each angle of \[{{P}_{1}}\] polygon to \[{{P}_{2}}\] polygon as 3:2.
We know that each angle of n sides of a regular polygon is given is as follows:
Each angle \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\]
So for \[{{P}_{1}}\] each angle \[=\left( \dfrac{{{N}_{1}}-2}{{{N}_{1}}} \right)\times {{180}^{\circ }}=\dfrac{2{{N}_{2}}-2}{2{{N}_{2}}}\times {{180}^{\circ }}\]
where \[{{N}_{1}}=2{{N}_{2}}\] from equation (1).
For \[{{P}_{2}}\] each angle \[\dfrac{{{N}_{2}}-2}{{{N}_{2}}}\times {{180}^{\circ }}\]
According to the question we have,
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{2{{N}_{2}}-2}{2{{N}_{2}}}\times {{180}}}{\dfrac{{{N}_{2}}-2}{{{N}_{2}}}\times 180}=\dfrac{3}{2} \\
 & \Rightarrow \dfrac{2\left( {{N}_{2}}-1 \right)\times {{N}_{2}}}{2{{N}_{2}}\times \left( {{N}_{2}}-2 \right)}=\dfrac{3}{2} \\
 & \Rightarrow \dfrac{{{N}_{2}}-1}{{{N}_{2}}-2}=\dfrac{3}{2} \\
\end{align}\]
On cross multiplication, we get as follows:
\[\begin{align}
  & \Rightarrow 2{{N}_{2}}-2=3{{N}_{2}}-6 \\
 & \Rightarrow 6-2=3{{N}_{2}}-2{{N}_{2}} \\
 & \Rightarrow 4={{N}_{2}} \\
\end{align}\]
Now substituting \[{{N}_{2}}=4\] in equation (4) we get as follows:
\[{{N}_{1}}=2\times 4=8\]
Hence \[{{N}_{1}}=8\] and \[{{N}_{2}}=8\].
Therefore the number of sides of two polygons are 4 and 8.

Note: Be careful while solving the equation and take care of the sign also. Don’t confuse anywhere of each angle of regular polygon \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\] where n is the number of sides of regular polygon properly. The ratio of angles that we take must be correct. If by mistake we take the ratio of angles of the second polygon to the first polygon as \[\dfrac{3}{2}\], we will end up with the wrong answer.