This question was asked in an exam of mathematics for which I am preparing and I was unable to solve it.
The ratio of number of sides of 2 regular polygons is 1:2. If each interior angle of 1st polygon is 120 then each interior angle of 2nd polygon is?
I can only think of that angle would be larger but I don't know any results which could be used( there is no specific syllabus for it in the exam).
This was asked in quiz (general concepts) section and I asked it on Puzzle stackexchange but It was closed there. So, I am asking for help here.
So, can you please help?
Each exterior angle of a regular polygon is $\frac{360º}{n}$.
Imagine a point tracing the sides of the polygon. The exterior angle is the angle turned after each side is traced. Since the point ends up where it started, it completes one full revolution or $360º$. Therefore, the sum of all the exterior angles (of any polygon, not just regular ones) is $360º$, and since all $n$ angles are the same for a regular polygon, each angle is $\frac{360º}{n}$.
Since the exterior angle and the interior angle lie on a straight line, the interior angle and the exterior angle sum to $180º$. Therefore, each interior angle is $180º - \frac{360º}{n}$ (which is equivalent to $\frac{180º(n-2)}{n}$ using algebra).
The number of sides of the 1st polygon is given by $180º - \frac{360º}{n} = 120º \Rightarrow n =6$. Guess and check works fine as well.
Therefore, the 2nd polygon has $12$ sides. Now you just have to plug in $n = 12$ into the interior angle formula.