# Thread: Angles between diagonals

1. ## Angles between diagonals

In a regular polygon there are two diagonals such that the angle between them is 50 degrees. What is the smallest number of sides of the polygon for which this is possible?

I had thought that for this to be the case, the smallest possible angle between diagonals of a polygon must be a factor of 50.

Factors of 50: 1, 50, 2, 25, 5, 10

Let x be the number of sides of the polygon
The biggest factor that lets x be a whole number is 10.
$\displaystyle \frac{360}{x}= 10$
$\displaystyle x=36$

WRONG!

2. Originally Posted by DivideBy0
In a regular polygon there are two diagonals such that the angle between them is 50 degrees. What is the smallest number of sides of the polygon for which this is possible?

I had thought that for this to be the case, the smallest possible angle between diagonals of a polygon must be a factor of 50.

Factors of 50: 1, 50, 2, 25, 5, 10

Let x be the number of sides of the polygon
The biggest factor that lets x be a whole number is 10.
$\displaystyle \frac{360}{x}= 10$
$\displaystyle x=36$

WRONG!

Hello,

you have considered the case that the diagonals intersect at the centre of the circle. But if the number of vertices is odd the diagonals don't intersect at the centre of the circle.

If you have a regular n-gon then you have:
n vertices
from one vertix are starting (n-1) lines (2 sides and (n-3) diagonals)
these lines are forming (n-2) angles which have the same size.
the size of these angles can be calculated by: $\displaystyle |angle| = \frac{360^\circ}{2n}$

In your case: $\displaystyle k \cdot \frac{360^\circ}{2n} = 50^\circ ~~\Longrightarrow ~~\frac{k}{n} = \frac{5}{18}$

That means: You have a 18-gon and between the diagonals in question there are 5 steps à 10°