# Thread: Equation needed - Regular polygon question!

1. ## Equation needed - Regular polygon question!

Question:
Each exterior angle of a regular polygon of n sides exceeds by 6deg each exterior ngle of a regular polygon of 2n sides. Find an equation for n and solve it.

I believe it involves some of the facts like:
* All sides and angles of a regular polygon are equal.
* The sum of exterior angles equals to 360deg.
* Each interior + exterior angle equals to 180deg.

But I am still not sure how to construct an equation for this problem. I guess I am just weak at constructing simultaneous or similar linear equations.

2. Hello, struck!

You have all the necessary information ... just put it together.

Each exterior angle of a regular polygon of n sides
exceeds by 6° each exterior angle of a regular polygon of 2n sides.
Find an equation for n and solve it.

I believe it involves some of the facts like:
. . All sides and angles of a regular polygon are equal.
. . The sum of exterior angles equals to 360°.
. . Each interior + exterior angle equals 180°.

The sum of the exterior angles is always 360°.

In a regular polygon of $\displaystyle n$ sides, each exterior angle has $\displaystyle \frac{360}{n}$ degrees. .[1]

In a regular polygon of $\displaystyle 2n$ sides, each exterior angle has $\displaystyle \frac{360}{2n}$ degrees. .[2]

We are told that [1] is 6° greater than [2].

Hence: . $\displaystyle \boxed{\frac{360}{n} \:=\:\frac{180}{n} + 6}$

Multiply by $\displaystyle n\!:\;\;360 \:=\:180 + 6n\quad\Rightarrow\quad\boxed{n\:=\:30}$