What is the formula for the interior angle of a regular polygon?

Each interior angle of a regular polygon with n sides is 180° − 360°/n, which is the same as (n − 2)·180°/n. For a pentagon it is 108°, for a hexagon 120°, for an octagon 135°.

How is the interior angle of a regular polygon derived?

The radii from the centre cut the polygon into n congruent isosceles triangles whose angles at the centre are 360°/n each. Their base angles are (180° − 360°/n)/2, and two of them meet at every vertex, so the interior angle is 180° − 360°/n.

What is the exterior angle of a regular polygon?

360°/n, the same as the angle at the centre between two neighbouring radii; the n exterior angles add up to 360°.

Angles of a Regular Polygon

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag $A$ to turn and resize the pentagon, or $O$ to move it; the readout under the canvas shows the angle at the centre, a base angle and an interior angle. Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Join the centre O to the five vertices. The five triangles OAB, OBC, … are congruent (SSS), so their angles at O are equal and fill a full turn: 360°/5 = 72° each. Each triangle is isosceles with two radii, so its base angles are (180° − 72°)/2 = 54°. At every vertex two base angles meet, so each interior angle is 2 · 54° = 108° = 180° − 360°/5. For n sides: 180° − 360°/n. Drag A to turn and resize the pentagon, or O to move it.Regular polygon angles

DisplayLabelsMeasures

Drag A to turn and resize the pentagon, or O to move it

∠AOB = 72°,  ∠OBA = 54°,  ∠ABC = 108°

Step 1 / 6

Step explanations

1A regular pentagon in its circle

Given
ABCDE is a regular pentagon: its five sides are equal (two ticks each), its five angles are equal, and its vertices lie on a circle with centre O. To prove: each interior angle is 180° − 360°/5 = 108°, and for a regular n-gon 180° − 360°/n.

Drag A: the pentagon turns and resizes. Learn more about the statement

Getting Started with the Visualizer

The Statement

Five Congruent Triangles

The Angle at the Centre

The Base Angles

Two Base Angles at Each Vertex

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

A

to turn and resize the pentagon, or

O

to move it, and watch the three angles in the readout stay at

72°

54°

and

108°

. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

Proof?Regular polygon anglesDisplayLabelsMeasuresDrag A to turn and resize the pentagon, or O to move it

∠AOB = 72°,  ∠OBA = 54°,  ∠ABC = 108°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1A regular pentagon in its circle

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — A turns and resizes the pentagon, O moves the circle; the hollow points B to E follow, 72° apart

4Readout — the line under the figure: the angle at the centre, a base angle and an interior angle, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 6 of 6

Proof?Regular polygon anglesDisplayLabelsMeasuresDrag A to turn and resize the pentagon, or O to move it

∠AOB = 72°,  ∠OBA = 54°,  ∠ABC = 108°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A regular pentagon in its circle

2Five congruent triangles

SSS
Join O to the five vertices. The five segments are radii, so they are equal (one tick each). Each of the triangles OAB, OBC, OCD, ODE, OEA has two radii and one side of the pentagon, so all five have the same three sides, and by SSS they are congruent.

3Each angle at O is 360°/5 = 72°

Angles around a point
The congruent triangles have equal angles at O, θ each, and the five angles θ fill the full turn round O: 5θ = 360°, so θ = 72°. For n sides, θ = 360°/n.

4The base angles are (180° − 72°)/2 = 54°

Isosceles base angles
The triangle OAB has two radii, OA = OB, so it is isosceles and its base angles at A and B are equal, φ each. Its angle sum gives θ + 2φ = 180°, so φ = (180° − 72°)/2 = 54°. The same holds in every one of the five triangles.

5∠ABC = 2φ = 108°

Adding
At the vertex B the radius OB splits the interior angle ABC into two parts: ∠ABO, a base angle of OAB, and ∠OBC, a base angle of OBC. Both are φ, so ∠ABC = 2φ = 108°. Every vertex is the same.

6Interior angle = 180° − 360°/n

Combining
In general 2φ = 180° − θ = 180° − 360°/n: 60° for the equilateral triangle, 90° for the square, 108° for the pentagon, 120° for the hexagon. As a check, n of them add up to n · 180° − 360° = (n − 2) · 180°, the polygon angle sum. ∎

12345

1The five radii — equal, one tick each, cutting the pentagon into five triangles

2The angles at the centre — five angles θ round O, 72° each

3The angle at B — β = φ + φ, made of the base angles of OAB and OBC

4Justification tags — Given, SSS, Angles around a point, Isosceles base angles, Adding, Combining

5Replay — at the last step Play becomes Replay and Next is off

The Statement

A regular polygon has all its sides equal and all its angles equal, and its vertices lie on a circle with centre

O

. The theorem gives the size of each interior angle of a regular polygon with

n

sides:

\text{interior angle} = 180° - \frac{360°}{n} = \frac{(n - 2) \cdot 180°}{n}

For the pentagon in the player that is

180° - 72° = 108°

. The two forms of the formula come from two proofs: the second divides the polygon angle sum by

n

, the first, proved here, works from the centre of the polygon, as on the regular hexagon page.

In the player

O

and

A

are free, and

B

E

are

A

turned about

O

72°

at a time; the readout shows the angle at the centre, a base angle and an interior angle.

Opening scene, frozen

The regular pentagon ABCDE in its circle, five equal sides.

Five Congruent Triangles

Join

O

to the five vertices. The five segments are radii, so they are equal. They cut the pentagon into five triangles

OAB

OBC

OCD

ODE

OEA

, each with two radii and one side of the pentagon. The sides are equal because the pentagon is regular, so the five triangles have the same three side lengths, and by SSS they are congruent.

Five congruent triangles, frozen

Five equal radii: five triangles congruent by SSS.

The Angle at the Centre

Congruent triangles have equal angles, so the five angles at

O

are equal,

\theta

each. Together they fill the full turn round

O

(angles around a point):

5\theta = 360°, \qquad \theta = 72°

For a regular

n

-gon,

\theta = 360°/n

; this is the central angle of the polygon.

The angle at the centre, frozen

Five equal angles θ round O: θ = 72°.

The Base Angles

Each of the five triangles is isosceles, with the two radii as equal sides, so its base angles are equal,

\varphi

each. The angle sum of the triangle gives

\theta + 2\varphi = 180°, \qquad \varphi = \frac{180° - 72°}{2} = 54°

The base angles, frozen

OAB shaded: base angles φ = (180° − 72°)/2 = 54°.

Two Base Angles at Each Vertex

At the vertex

B

the radius

OB

splits the interior angle

ABC

into two parts. One part,

\angle ABO

, is a base angle of the triangle

OAB

; the other,

\angle OBC

, is a base angle of

OBC

. Both are

\varphi

, so

\angle ABC = 2\varphi = 108°

and the same at every vertex, since all five triangles are congruent.

Two base angles at each vertex, frozen

The angle at B is φ + φ = 108°.

Why It Holds

In general

2\varphi = 180° - \theta

, so

\text{interior angle} = 180° - \frac{360°}{n}

60°

for the equilateral triangle,

90°

for the square,

108°

for the pentagon,

120°

for the hexagon,

135°

for the octagon. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, SSS, angles around a point, isosceles base angles, adding, and combining.

As a check,

n

such angles add up to

n \cdot 180° - 360° = (n - 2) \cdot 180°

, the polygon angle sum. And the exterior angle at each vertex is

180°

minus the interior one,

360°/n

, the same as the angle at the centre;

n

of them make the

360°

of the polygon exterior angles page.

The conclusion, frozen

Each interior angle is 180° − 360°/n.

Dragging the Figure

Drag

A

: the whole pentagon turns with it and grows or shrinks, since

B

E

are always

A

turned about

O

72°

steps, and the three angles in the readout do not change. Drag

O

to move the circle;

A

stays where it is, so the pentagon turns and resizes about the new centre.

The player keeps the radius within a range that keeps the five angle marks at the centre readable and the pentagon in the frame; a drag outside that range is refused.

After dragging A, frozen

A moved: the pentagon turned and grew, every angle still 108°.

Consequences

• Tilings. Copies of one regular polygon fill the plane round a vertex only if the interior angle divides

360°

60°

90°

and

120°

do,

108°

does not, so the only regular tilings use triangles, squares and hexagons.
• Exterior angle = central angle. Both are

360°/n

: turning by the exterior angle at each vertex, a walker turns as far as the radius does from one vertex to the next.
• Counting sides from the angle. An interior angle of

140°

means

360°/n = 40°

, so

n = 9

; any angle

180° - 360°/n

fixes

n

.
• The hexagon. For

n = 6

the base angles are

60°

, so the triangles are equilateral and the side equals the radius, the regular hexagon theorem.
• Diagonals of the pentagon. The angles

108°

and

36°

that appear when the diagonals are drawn give the golden ratio as the ratio of diagonal to side, through similar triangles.