Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need a theoretical proof that each angle of a regular hexagon is $120^\circ$.

share|improve this question
add comment

3 Answers

Draw the five radiuses from the hexagonal's center to its vertices. You get 6 congruent isosceles triangles whose basis angle's equals $\,x=\,$ half our wanted angle..

But then the central angle in each triangle equals $\,180^\circ-2x\,$ , so

$$6(180^\circ-2x)=360^\circ\Longrightarrow 180^\circ-2x=60^\circ\Longrightarrow 2x=120^\circ$$

share|improve this answer
add comment

Sum of the internal angles of a regular Hexagon is $(2.6-4)*90=720 $ As there are 6 angles so each angle must be $720/6=120$

Another Perspective:In a regular Hexagon there are 6 Triangles each having the sum of the angles as $180$ degrees.So the sum of the Angles of all the 6 triangles is $180*6$ but this contains the central angle of $360$ degree so the sum of the internal angles of the hexagon is $180.6-360=720$.As all the angles are same so each angle must be $120$ degrees.

share|improve this answer
add comment

Consider a line rotating about a regular $n$-gon in the Euclidean plane. If the angle at each vertex is $a$, then the line rotates $180-a$ at each vertex.

Since there are $n$ vertices, the total rotation is $n(180-a)$.

But this total rotation is $360$. So, $n(180-a) = 360$ or $a = 180 - 360/n$.

For $n = 6$, $a = 180-360/6 = 180-60 = 120$.

Note: This is not original by me.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.