Exterior Angles and Regular Polygons

Find the exterior angles of a polygon and see why they always add up to 360°. Use this to find each angle of a regular polygon, its number of sides, and how regular polygons fit together at a point.

Before you start

A polygon is convex if every interior angle is less than 180° (the lesson Polygons and Regular Polygons). The interior angles of a convex polygon with $n$ sides add up to $(n - 2) \times 180^{\circ}$ , written (∠ sum of polygon), short for "angle sum of a polygon". For a hexagon, $n = 6$ and the sum is $4 \times 180^{\circ} = 720^{\circ}$ (the lesson Polygons and Their Interior Angles).
A regular polygon has all its sides equal and all its angles equal. A regular polygon with $n$ sides has $n$ lines of symmetry and rotational symmetry of order $n$ (the lesson Polygons and Regular Polygons). Each of its interior angles is $\frac{(n - 2) \times 180^{\circ}}{n}$ : for a regular hexagon, $720^{\circ} \div 6 = 120^{\circ}$ (the lesson Polygons and Their Interior Angles).
An exterior angle of a triangle is the angle between one side and the extension of an adjacent side. It and the interior angle beside it add up to 180° (adj. ∠s on a str. line) (the lesson Triangles and Their Angles).
Angles at a point add up to 360° (∠s at a pt) (Chapter 11).

The Exterior Angle of a Polygon

In the lesson Triangles and Their Angles you met the exterior angle of a triangle. Every convex polygon has exterior angles in the same way: extend one side past a vertex, and the angle between the extension and the other side at that vertex is an exterior angle there.

In the figure, side $AB$ of pentagon $ABCDE$ is extended past $B$ to $F$ . The exterior angle at $B$ is $\angle CBF$ , between the side $BC$ and the extension $BF$ . The interior angle at $B$ is $\angle ABC$ .

$\angle ABC$ and $\angle CBF$ lie together on the straight line $ABF$ , so they add up to 180°: $115^{\circ} + 65^{\circ} = 180^{\circ}$ .

The exterior angle of a convex polygon

An exterior angle of a convex polygon is the angle between one side and the extension of an adjacent side, at the vertex where they meet.

At each vertex, interior angle + exterior angle $= 180^{\circ}$ (adj. ∠s on a str. line).

To mark the exterior angles of the whole polygon, go round it in one direction and extend each side the same way round: $AB$ past $B$ , $BC$ past $C$ , $CD$ past $D$ , $DE$ past $E$ and $EA$ past $A$ . That gives one exterior angle at each vertex, marked $a$ , $b$ , $c$ , $d$ and $e$ below.

One exterior angle at each vertex

At $B$ you could extend $CB$ past $B$ instead of $AB$ . The angle you get is vertically opposite $\angle CBF$ , so it is the same size (vert. opp. ∠s). Either one is the exterior angle at $B$ : when you add up the exterior angles of a polygon, take one at each vertex.

The reflex angle all the way round the outside of $B$ , $360^{\circ} - 115^{\circ} = 245^{\circ}$ , is not an exterior angle.

Example 1: An exterior angle and an interior angle

In pentagon $PQRST$ , $PQ$ is extended to $U$ and $RS$ is extended to $V$ . $\angle PQR = 118^{\circ}$ and $\angle TSV = 64^{\circ}$ . Find $x$ and $y$ .

Solution

$\angle PQR$ and $x^{\circ}$ lie together on the straight line $PQU$ :

$x = 180 - 118 = 62$ (adj. ∠s on a str. line)

$y^{\circ}$ and $\angle TSV$ lie together on the straight line $RSV$ :

$y = 180 - 64 = 116$ (adj. ∠s on a str. line)

Check: $118^{\circ} + 62^{\circ} = 180^{\circ}$ and $116^{\circ} + 64^{\circ} = 180^{\circ}$ .

$x = 62$ and $y = 116$ : the exterior angle at $Q$ is 62° and the interior angle at $S$ is 116°.

The Sum of the Exterior Angles

Look again at pentagon $ABCDE$ with its exterior angles $a$ , $b$ , $c$ , $d$ and $e$ , drawn earlier in this lesson. Imagine walking round it. Start at $A$ , facing along $AB$ , and walk to $B$ . At $B$ you turn left through the exterior angle $b$ to face along $BC$ . Walk to $C$ and turn through $c$ ; at $D$ turn through $d$ ; at $E$ turn through $e$ ; and back at $A$ turn through $a$ . You are now facing along $AB$ again, the way you started.

On the way round you have turned through exactly one complete turn, so $a + b + c + d + e = 360^{\circ}$ .

You can see the complete turn in another way. From one point $O$ , draw a line in the direction of each side, in order. The angle from each line to the next is the turn at a vertex, an exterior angle, and together the five exterior angles fill the whole of the angle round $O$ (∠s at a pt).

The same is true of every convex polygon, and you can also show it from the interior angles. A polygon with $n$ sides has $n$ vertices. At each one, interior angle + exterior angle $= 180^{\circ}$ , so all the interior and exterior angles together make $n \times 180^{\circ}$ . The interior angles alone make $(n - 2) \times 180^{\circ}$ (∠ sum of polygon). So the exterior angles make

$n \times 180^{\circ} - (n - 2) \times 180^{\circ} = n \times 180^{\circ} - n \times 180^{\circ} + 2 \times 180^{\circ} = 360^{\circ}$ .

For pentagon $ABCDE$ : $5 \times 180^{\circ} - 3 \times 180^{\circ} = 900^{\circ} - 540^{\circ} = 360^{\circ}$ .

The sum of the exterior angles of a polygon

The exterior angles of any convex polygon, one at each vertex, add up to 360°.

This does not depend on the number of sides: the exterior angles of a triangle, of a pentagon and of a polygon with 20 sides all add up to 360°.

The reason is written (sum of ext. ∠s of polygon), short for "the sum of the exterior angles of a polygon is 360°".

Example 2: The missing exterior angle

The figure shows hexagon $ABCDEF$ with each side extended the same way round. Five of its exterior angles are 50°, 65°, 70°, 48° and 55°. Find $x$ .

Solution

The six exterior angles, one at each vertex, add up to 360°:

$x + 50 + 65 + 70 + 48 + 55 = 360$ (sum of ext. ∠s of polygon)

$x + 288 = 360$

$x = 72$

Check: $72^{\circ} + 50^{\circ} + 65^{\circ} + 70^{\circ} + 48^{\circ} + 55^{\circ} = 360^{\circ}$ .

$x = 72$ , so the exterior angle at $F$ is 72°.

Example 3: Exterior angles in x

The exterior angles of pentagon $ABCDE$ are $2x^{\circ}$ , $(x + 15)^{\circ}$ , $(3x - 20)^{\circ}$ , $(x + 5)^{\circ}$ and $2x^{\circ}$ , as in the figure. Find $x$ , and hence find the interior angle $\angle BCD$ .

Solution

$2x + (x + 15) + (3x - 20) + (x + 5) + 2x = 360$ (sum of ext. ∠s of polygon)

Collect like terms: $9x + 15 - 20 + 5 = 360$ , so $9x = 360$ .

Divide both sides by 9: $x = 40$

The exterior angles are $80^{\circ}$ , $55^{\circ}$ , $100^{\circ}$ , $45^{\circ}$ and $80^{\circ}$ . Check: $80 + 55 + 100 + 45 + 80 = 360$ .

The exterior angle at $C$ is $(3 \times 40 - 20)^{\circ} = 100^{\circ}$ , so

$\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}$ (adj. ∠s on a str. line)

$x = 40$ and $\angle BCD = 80^{\circ}$ .

Example 4: Interior and exterior angles together

In pentagon $ABCDE$ , $\angle EAB = 100^{\circ}$ , $\angle ABC = 120^{\circ}$ and $\angle BCD = 110^{\circ}$ . $CD$ is extended to $F$ , and $\angle EDF = 75^{\circ}$ . Find $x$ .

Solution

Find the exterior angles at $A$ , $B$ and $C$ (adj. ∠s on a str. line):

at $A$ : $180^{\circ} - 100^{\circ} = 80^{\circ}$ ; at $B$ : $180^{\circ} - 120^{\circ} = 60^{\circ}$ ; at $C$ : $180^{\circ} - 110^{\circ} = 70^{\circ}$ .

The exterior angle at $D$ is $\angle EDF = 75^{\circ}$ . So the exterior angle at $E$ is

$360^{\circ} - 80^{\circ} - 60^{\circ} - 70^{\circ} - 75^{\circ} = 75^{\circ}$ (sum of ext. ∠s of polygon)

$x = 180 - 75 = 105$ (adj. ∠s on a str. line)

Check with the interior angles: $\angle CDE = 180^{\circ} - 75^{\circ} = 105^{\circ}$ , and $100^{\circ} + 120^{\circ} + 110^{\circ} + 105^{\circ} + 105^{\circ} = 540^{\circ} = (5 - 2) \times 180^{\circ}$ (∠ sum of polygon).

$x = 105$ , so $\angle AED = 105^{\circ}$ .

Each Angle of a Regular Polygon

In a regular polygon all the interior angles are equal. Each exterior angle is 180° minus an interior angle, so all the exterior angles are equal too.

A regular polygon with $n$ sides has $n$ equal exterior angles, and they add up to 360°. So each exterior angle is $360^{\circ} \div n$ . For a regular pentagon, each exterior angle is $360^{\circ} \div 5 = 72^{\circ}$ , and each interior angle is $180^{\circ} - 72^{\circ} = 108^{\circ}$ .

The angles of a regular polygon

For a regular polygon with $n$ sides:

each exterior angle $= \frac{360^{\circ}}{n}$ (ext. ∠ of regular polygon);

each interior angle $= 180^{\circ} -$ each exterior angle (adj. ∠s on a str. line);

the number of sides $n = \frac{360^{\circ}}{\text{each exterior angle}}$ .

The reason (ext. ∠ of regular polygon) is short for "each exterior angle of a regular polygon with $n$ sides is $\frac{360^{\circ}}{n}$ ".

This is a quicker way to each interior angle than the angle sum. For a regular octagon, each exterior angle is $360^{\circ} \div 8 = 45^{\circ}$ , so each interior angle is $180^{\circ} - 45^{\circ} = 135^{\circ}$ . From the angle sum, $\frac{(8 - 2) \times 180^{\circ}}{8} = \frac{1080^{\circ}}{8} = 135^{\circ}$ : the same.

The four regular polygons you will meet most often are drawn below, each with one interior angle (blue) and one exterior angle (amber) marked.

Regular pentagon

Regular hexagon

Regular octagon

Regular decagon

Regular polygon	Number of sides	Each exterior angle	Each interior angle	Lines of symmetry	Order of rotational symmetry
Pentagon	5	$360^{\circ} \div 5 = 72^{\circ}$	$108^{\circ}$	5	5
Hexagon	6	$360^{\circ} \div 6 = 60^{\circ}$	$120^{\circ}$	6	6
Octagon	8	$360^{\circ} \div 8 = 45^{\circ}$	$135^{\circ}$	8	8
Decagon	10	$360^{\circ} \div 10 = 36^{\circ}$	$144^{\circ}$	10	10

Example 5: The angles of a regular polygon

A regular polygon has 40 sides. Find (a) each exterior angle, (b) each interior angle.

Solution

(a) Each exterior angle $= \frac{360^{\circ}}{40} = 9^{\circ}$ (ext. ∠ of regular polygon)

(b) Each interior angle $= 180^{\circ} - 9^{\circ} = 171^{\circ}$ (adj. ∠s on a str. line)

Check with the angle sum: $(40 - 2) \times 180^{\circ} = 6840^{\circ}$ , and $6840^{\circ} \div 40 = 171^{\circ}$ .

(a) Each exterior angle is 9°. (b) Each interior angle is 171°.

Finding the Number of Sides

Turn the rule round: if you know each exterior angle of a regular polygon, divide 360° by it to find the number of sides. If you know each interior angle, find the exterior angle first.

Example 6: The number of sides from an exterior angle

Each exterior angle of a regular polygon is 20°. Find (a) the number of sides, (b) each interior angle.

Solution

(a) $n = \frac{360^{\circ}}{20^{\circ}} = 18$ (ext. ∠ of regular polygon)

(b) Each interior angle $= 180^{\circ} - 20^{\circ} = 160^{\circ}$ (adj. ∠s on a str. line)

Check: $(18 - 2) \times 180^{\circ} = 2880^{\circ}$ , and $2880^{\circ} \div 18 = 160^{\circ}$ .

(a) The polygon has 18 sides. (b) Each interior angle is 160°.

Example 7: The number of sides from an interior angle

Each interior angle of a regular polygon is 162°. How many sides has the polygon?

Solution

Each exterior angle $= 180^{\circ} - 162^{\circ} = 18^{\circ}$ (adj. ∠s on a str. line)

$n = \frac{360^{\circ}}{18^{\circ}} = 20$ (ext. ∠ of regular polygon)

Check: $(20 - 2) \times 180^{\circ} = 3240^{\circ}$ , and $3240^{\circ} \div 20 = 162^{\circ}$ .

Do not divide 360° by the interior angle: $360^{\circ} \div 162^{\circ}$ is not the number of sides.

The polygon has 20 sides.

Not every angle can be an angle of a regular polygon. The number of sides must be a whole number, and a polygon has at least 3 sides. So 360° divided by each exterior angle must be a whole number, 3 or more.

The fewer the sides, the larger each exterior angle. The largest belongs to the equilateral triangle: $360^{\circ} \div 3 = 120^{\circ}$ . So each exterior angle of a regular polygon is at most 120°, and each interior angle is at least $180^{\circ} - 120^{\circ} = 60^{\circ}$ . Each exterior angle is more than 0°, so each interior angle is less than 180°.

Which angles can belong to a regular polygon?

An angle can be each exterior angle of a regular polygon exactly when $360^{\circ} \div$ that angle is a whole number, 3 or more; that whole number is the number of sides.

Each exterior angle of a regular polygon is at most 120°. Each interior angle is at least 60° and less than 180°.

To test an interior angle, find the exterior angle first, then divide 360° by it.

Example 8: Can it be a regular polygon?

Can each interior angle of a regular polygon be (a) 130°, (b) 165°? If it can, find the number of sides.

Solution

(a) Each exterior angle would be $180^{\circ} - 130^{\circ} = 50^{\circ}$ , and $360^{\circ} \div 50^{\circ} = 7.2$ .

The number of sides would not be a whole number, so no regular polygon has interior angles of 130°. (A regular polygon with 7 sides has exterior angles of about 51.4°, and one with 8 sides has exterior angles of 45°: 50° falls between them.)

(b) Each exterior angle would be $180^{\circ} - 165^{\circ} = 15^{\circ}$ , and $360^{\circ} \div 15^{\circ} = 24$ , a whole number. So it can: the polygon has 24 sides.

(c) $360^{\circ} \div 150^{\circ} = 2.4$ , not a whole number; and 150° is more than 120°, the largest exterior angle of any regular polygon. So it cannot.

(a) No: it would need 7.2 sides. (b) Yes: 24 sides. (c) No: an exterior angle of a regular polygon is at most 120°.

Example 9: The interior angle is 3 times the exterior angle

Each interior angle of a regular polygon is 3 times each exterior angle. Find each exterior angle and the number of sides, and name the polygon.

Solution

Let each exterior angle be $x^{\circ}$ . Then each interior angle is $3x^{\circ}$ , and the two add up to 180°:

$x + 3x = 180$ (adj. ∠s on a str. line)

$4x = 180$ , so $x = 45$ .

$n = \frac{360^{\circ}}{45^{\circ}} = 8$ (ext. ∠ of regular polygon)

Check: each interior angle is $180^{\circ} - 45^{\circ} = 135^{\circ}$ , and $135^{\circ} = 3 \times 45^{\circ}$ .

Each exterior angle is 45°; the polygon has 8 sides: it is a regular octagon.

Regular Polygons That Meet

Where regular polygons meet at a point, all the angles round that point, including any gap, add up to 360° (∠s at a pt). Where they share a side, their sides are equal, which often gives an isosceles triangle.

Example 10: A square on the side of a regular hexagon

In the figure, $ABCDEF$ is a regular hexagon and $ABGH$ is a square drawn on the side $AB$ , outside the hexagon. $F$ is joined to $H$ . Find $x$ and $y$ .

Solution

Each exterior angle of a regular hexagon is $360^{\circ} \div 6 = 60^{\circ}$ (ext. ∠ of regular polygon), so each interior angle is $180^{\circ} - 60^{\circ} = 120^{\circ}$ (adj. ∠s on a str. line). So $\angle FAB = 120^{\circ}$ .

$\angle BAH = 90^{\circ}$ , an angle of the square.

The three angles at $A$ make a complete turn:

$x + 120 + 90 = 360$ (∠s at a pt)

$x = 150$

The hexagon and the square share the side $AB$ , so $AF = AB$ (sides of the regular hexagon) and $AH = AB$ (sides of the square). So $AF = AH$ , and $\triangle AFH$ is isosceles with apex $A$ :

$\angle AFH = \angle AHF$ (base ∠s of isos. △), so $y = (180 - 150) \div 2 = 15$ (∠ sum of △)

Check in $\triangle AFH$ : $150^{\circ} + 15^{\circ} + 15^{\circ} = 180^{\circ}$ .

$x = 150$ and $y = 15$ : $\angle FAH = 150^{\circ}$ and $\angle AFH = 15^{\circ}$ .

Example 11: Why hexagons fit together and pentagons do not

The cells of a honeycomb are regular hexagons that fit together with no gaps. Mr Tan wants to cover a floor with tiles that are all the same regular polygon, meeting corner to corner with no gaps and no overlaps. Show that regular hexagons fit together round a point, and that regular pentagons do not.

Solution

Where tiles meet at a point, their angles at that point must add up to exactly 360° (∠s at a pt).

Each exterior angle of a regular hexagon is $360^{\circ} \div 6 = 60^{\circ}$ , so each interior angle is $180^{\circ} - 60^{\circ} = 120^{\circ}$ . Three hexagons give $3 \times 120^{\circ} = 360^{\circ}$ : they fit exactly round the point.

Each exterior angle of a regular pentagon is $360^{\circ} \div 5 = 72^{\circ}$ , so each interior angle is $180^{\circ} - 72^{\circ} = 108^{\circ}$ . Three pentagons give $3 \times 108^{\circ} = 324^{\circ}$ , which leaves a gap of $360^{\circ} - 324^{\circ} = 36^{\circ}$ ; four would need $4 \times 108^{\circ} = 432^{\circ}$ , more than 360°, so they would overlap.

Hexagons:

3 \times 120^{\circ} = 360^{\circ}

Pentagons: a gap of

36^{\circ}

In the same way, copies of one regular polygon fit exactly round a point only if its interior angle divides 360° exactly: the equilateral triangle (6 × 60°), the square (4 × 90°) and the regular hexagon (3 × 120°). Every other regular polygon has an interior angle of 108° (the pentagon) or between 120° and 180° (7 sides or more), and 360° divided by any of these is not a whole number.

Three regular hexagons make $3 \times 120^{\circ} = 360^{\circ}$ and fit exactly; $360^{\circ} \div 108^{\circ}$ is not a whole number, so regular pentagons leave a gap or overlap.

Common mistakes

Taking the exterior angle as $360^{\circ} -$ the interior angle. That is the reflex angle outside the vertex. The exterior angle is $180^{\circ} -$ the interior angle.
Using $(n - 2) \times 180^{\circ}$ for the exterior angles. That is the sum of the interior angles. The exterior angles always add up to 360°, whatever the number of sides.
Dividing 360° by an interior angle to find the number of sides. For a regular hexagon, $360^{\circ} \div 120^{\circ} = 3$ , not 6. Find the exterior angle first: $360^{\circ} \div 60^{\circ} = 6$ .
Counting two exterior angles at the same vertex, or leaving a vertex out. The sum of 360° takes exactly one exterior angle at each vertex.
Saying an angle belongs to a regular polygon without checking that the number of sides is a whole number, 3 or more. An interior angle of 130° gives $360^{\circ} \div 50^{\circ} = 7.2$ : no regular polygon has it.

Practice

Try each question before you open its solution. Give a reason for each step.

Practice 1: The fourth exterior angle

The figure shows quadrilateral $ABCD$ with each side extended the same way round. Its exterior angles at $A$ , $B$ and $C$ are 100°, 85° and 105°.

Find (a) $x$ , (b) the interior angle $\angle ADC$ .

ShowHide solution

(a) $x + 100 + 85 + 105 = 360$ (sum of ext. ∠s of polygon)

$x + 290 = 360$ , so $x = 70$ .

(b) $\angle ADC = 180^{\circ} - 70^{\circ} = 110^{\circ}$ (adj. ∠s on a str. line)

Check with the interior angles: they are $180^{\circ} - 100^{\circ} = 80^{\circ}$ , $180^{\circ} - 85^{\circ} = 95^{\circ}$ , $180^{\circ} - 105^{\circ} = 75^{\circ}$ and $110^{\circ}$ , and $80^{\circ} + 95^{\circ} + 75^{\circ} + 110^{\circ} = 360^{\circ} = (4 - 2) \times 180^{\circ}$ (∠ sum of polygon).

(a) $x = 70$ (b) $\angle ADC = 110^{\circ}$ .

Practice 2: A regular polygon with 30 sides

A regular polygon has 30 sides. Find (a) each exterior angle, (b) each interior angle.

ShowHide solution

(a) Each exterior angle $= \frac{360^{\circ}}{30} = 12^{\circ}$ (ext. ∠ of regular polygon)

(b) Each interior angle $= 180^{\circ} - 12^{\circ} = 168^{\circ}$ (adj. ∠s on a str. line)

Check: $(30 - 2) \times 180^{\circ} = 5040^{\circ}$ , and $5040^{\circ} \div 30 = 168^{\circ}$ .

(a) 12° (b) 168°.

Practice 3: Jogging round a path

Ali jogs round a path in the park in the shape of a regular pentagon. At each corner he turns left to run along the next side.

(a) Through what angle does he turn at each corner?

(b) Through what angle has he turned altogether when he is back at his starting corner, facing the way he started?

ShowHide solution

(a) At each corner he turns through the exterior angle: $360^{\circ} \div 5 = 72^{\circ}$ (ext. ∠ of regular polygon).

(b) He turns at 5 corners: $5 \times 72^{\circ} = 360^{\circ}$ , one complete turn (sum of ext. ∠s of polygon).

(a) 72° (b) 360°, one complete turn (c) 108°.

Practice 4: From the exterior angle

Each exterior angle of a regular polygon is 10°. Find (a) the number of sides, (b) the sum of its interior angles.

ShowHide solution

(a) $n = \frac{360^{\circ}}{10^{\circ}} = 36$ (ext. ∠ of regular polygon)

(b) Sum of the interior angles $= (36 - 2) \times 180^{\circ} = 34 \times 180^{\circ} = 6120^{\circ}$ (∠ sum of polygon)

Check: each interior angle is $180^{\circ} - 10^{\circ} = 170^{\circ}$ , and $36 \times 170^{\circ} = 6120^{\circ}$ .

(a) 36 sides (b) 6120°.

Practice 5: From the interior angle

Each interior angle of a regular polygon is 156°. Find the number of sides.

ShowHide solution

Each exterior angle $= 180^{\circ} - 156^{\circ} = 24^{\circ}$ (adj. ∠s on a str. line)

$n = \frac{360^{\circ}}{24^{\circ}} = 15$ (ext. ∠ of regular polygon)

Check: $(15 - 2) \times 180^{\circ} = 2340^{\circ}$ , and $2340^{\circ} \div 15 = 156^{\circ}$ .

The polygon has 15 sides.

Practice 6: Six exterior angles in x

The figure shows hexagon $ABCDEF$ with each side extended the same way round.

Find $x$ , and hence the smallest interior angle of the hexagon.

ShowHide solution

$2x + (x + 30) + (3x - 40) + (x + 10) + (2x - 20) + x = 360$ (sum of ext. ∠s of polygon)

Collect like terms: $10x - 20 = 360$

Add 20 to both sides: $10x = 380$

Divide both sides by 10: $x = 38$

The exterior angles are $76^{\circ}$ , $68^{\circ}$ , $74^{\circ}$ , $48^{\circ}$ , $56^{\circ}$ and $38^{\circ}$ . Check: $76 + 68 + 74 + 48 + 56 + 38 = 360$ .

The smallest interior angle is beside the largest exterior angle, $76^{\circ}$ at $A$ :

$\angle FAB = 180^{\circ} - 76^{\circ} = 104^{\circ}$ (adj. ∠s on a str. line)

Check: the interior angles are $104^{\circ}$ , $112^{\circ}$ , $106^{\circ}$ , $132^{\circ}$ , $124^{\circ}$ and $142^{\circ}$ , which add up to $720^{\circ} = (6 - 2) \times 180^{\circ}$ (∠ sum of polygon).

$x = 38$ ; the smallest interior angle is $\angle FAB = 104^{\circ}$ .

Practice 7: Some interior angles given

In pentagon $PQRST$ , $\angle TPQ = 95^{\circ}$ , $\angle PQR = 110^{\circ}$ and $\angle QRS = 125^{\circ}$ . $RS$ is extended to $U$ , and $\angle TSU = 80^{\circ}$ . Find $x$ .

ShowHide solution

Find the exterior angles at $P$ , $Q$ and $R$ (adj. ∠s on a str. line):

at $P$ : $180^{\circ} - 95^{\circ} = 85^{\circ}$ ; at $Q$ : $180^{\circ} - 110^{\circ} = 70^{\circ}$ ; at $R$ : $180^{\circ} - 125^{\circ} = 55^{\circ}$ .

The exterior angle at $S$ is $\angle TSU = 80^{\circ}$ . So the exterior angle at $T$ is

$360^{\circ} - 85^{\circ} - 70^{\circ} - 55^{\circ} - 80^{\circ} = 70^{\circ}$ (sum of ext. ∠s of polygon)

$x = 180 - 70 = 110$ (adj. ∠s on a str. line)

Check with the interior angles: $\angle RST = 180^{\circ} - 80^{\circ} = 100^{\circ}$ , and $95^{\circ} + 110^{\circ} + 125^{\circ} + 100^{\circ} + 110^{\circ} = 540^{\circ} = (5 - 2) \times 180^{\circ}$ (∠ sum of polygon).

$x = 110$ , so $\angle PTS = 110^{\circ}$ .

Practice 8: Which angles are possible?

Which of these can be each interior angle of a regular polygon? For each one that can, give the number of sides.

(a) 172° (b) 145° (c) 175°

(d) Priya says a regular polygon can have each exterior angle 100°, because 100° is less than 120°. Is she right? Explain.

ShowHide solution

(a) Exterior angle $180^{\circ} - 172^{\circ} = 8^{\circ}$ ; $360^{\circ} \div 8^{\circ} = 45$ , a whole number. Yes: 45 sides.

(b) Exterior angle $180^{\circ} - 145^{\circ} = 35^{\circ}$ ; $360^{\circ} \div 35^{\circ}$ is between 10 and 11 ( $35 \times 10 = 350$ and $35 \times 11 = 385$ ), not a whole number. No.

(c) Exterior angle $180^{\circ} - 175^{\circ} = 5^{\circ}$ ; $360^{\circ} \div 5^{\circ} = 72$ , a whole number. Yes: 72 sides.

(d) $360^{\circ} \div 100^{\circ} = 3.6$ , not a whole number, so no regular polygon has exterior angles of 100°. (A regular polygon with 3 sides has exterior angles of 120°, and one with 4 sides has exterior angles of 90°: 100° falls between them.) Being at most 120° is not enough: $360^{\circ} \div$ the angle must also be a whole number.

(a) Yes, 45 sides (b) No (c) Yes, 72 sides (d) No: 360° ÷ 100° = 3.6 is not a whole number. Being at most 120° is not enough.

Practice 9: The interior angle is 108° more

Each interior angle of a regular polygon is 108° more than each exterior angle. Find each exterior angle, and name the polygon.

ShowHide solution

Let each exterior angle be $x^{\circ}$ . Then each interior angle is $(x + 108)^{\circ}$ .

$x + (x + 108) = 180$ (adj. ∠s on a str. line)

$2x + 108 = 180$ , so $2x = 72$ and $x = 36$ .

$n = \frac{360^{\circ}}{36^{\circ}} = 10$ (ext. ∠ of regular polygon)

Check: each interior angle is $180^{\circ} - 36^{\circ} = 144^{\circ}$ , and $144^{\circ} - 36^{\circ} = 108^{\circ}$ .

Each exterior angle is 36°; the polygon has 10 sides: it is a regular decagon.

Practice 10: A robot on the floor

Wei Ming programs a robot on the classroom floor. It moves forward 50 cm, then turns left through 40°, and repeats these two steps until it is back where it started, facing the way it started.

(a) How many times does it turn?

(b) Describe the shape of its path.

ShowHide solution

(a) Each turn is an exterior angle of the path, and the turns add up to one complete turn, 360° (sum of ext. ∠s of polygon). So it turns $360^{\circ} \div 40^{\circ} = 9$ times.

(b) The path has 9 sides, each 50 cm long, and 9 equal angles: it is a regular polygon with 9 sides (a regular nonagon).

Check: $(9 - 2) \times 180^{\circ} = 1260^{\circ}$ , and $1260^{\circ} \div 9 = 140^{\circ}$ .

(a) 9 times (b) a regular polygon with 9 sides of 50 cm (a regular nonagon) (c) 140°.

Practice 11: Octagon and square floor tiles

Mdm Lim's kitchen floor is covered with tiles in the shapes of regular octagons and squares, all with sides of the same length. At each point where tiles meet, two octagons and one square meet, as at $O$ in the figure.

(a) Find each interior angle of a regular octagon.

(b) Show that two octagons and a square fit exactly round $O$ .

ShowHide solution

(a) Each exterior angle $= 360^{\circ} \div 8 = 45^{\circ}$ (ext. ∠ of regular polygon), so each interior angle $= 180^{\circ} - 45^{\circ} = 135^{\circ}$ (adj. ∠s on a str. line).

(b) The angles at $O$ : $135^{\circ} + 135^{\circ} + 90^{\circ} = 360^{\circ}$ , exactly the angle round a point (∠s at a pt). So they fit with no gap and no overlap.

(c) Two octagons make $270^{\circ}$ , leaving a gap of $90^{\circ}$ ; three would need $405^{\circ}$ , more than 360°. Since $360^{\circ} \div 135^{\circ}$ is not a whole number, octagons alone always leave gaps; the square tiles fill them.

(a) 135° (b) $135^{\circ} + 135^{\circ} + 90^{\circ} = 360^{\circ}$ (c) 360° ÷ 135° is not a whole number, so octagons alone leave gaps.

Practice 12: A triangle inside a regular pentagon

$ABCDE$ is a regular pentagon. An equilateral triangle $ABF$ is drawn on the side $AB$ , inside the pentagon, and $E$ is joined to $F$ .

Find (a) $\angle EAF$ , (b) $\angle AEF$ , (c) $\angle DEF$ .

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(a) Each interior angle of a regular pentagon is $180^{\circ} - 360^{\circ} \div 5 = 180^{\circ} - 72^{\circ} = 108^{\circ}$ (ext. ∠ of regular polygon; adj. ∠s on a str. line), so $\angle EAB = 108^{\circ}$ .

$\angle FAB = 60^{\circ}$ (equilateral △)

$\angle EAF = 108^{\circ} - 60^{\circ} = 48^{\circ}$

(b) $AE = AB$ (sides of the regular pentagon) and $AF = AB$ (sides of the equilateral triangle), so $AE = AF$ and $\triangle AEF$ is isosceles with apex $A$ :

$\angle AEF = \angle AFE$ (base ∠s of isos. △), so $\angle AEF = (180^{\circ} - 48^{\circ}) \div 2 = 66^{\circ}$ (∠ sum of △)

Check in $\triangle AEF$ : $48^{\circ} + 66^{\circ} + 66^{\circ} = 180^{\circ}$ .

(a) $\angle EAF = 48^{\circ}$ (b) $\angle AEF = 66^{\circ}$ (c) $\angle DEF = 42^{\circ}$ .

Practice 13: The third tile

A square tile and a regular hexagon tile, with sides of the same length, meet at a point $O$ and share a side, as in the figure. A third tile, also a regular polygon with sides of the same length, fills the angle $x^{\circ}$ left at $O$ exactly. How many sides has the third tile?

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Each exterior angle of a regular hexagon is $360^{\circ} \div 6 = 60^{\circ}$ , so its interior angle at $O$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$ . The square's angle at $O$ is $90^{\circ}$ .

$x + 90 + 120 = 360$ (∠s at a pt)

$x = 150$

So each interior angle of the third tile is $150^{\circ}$ , and each exterior angle is $180^{\circ} - 150^{\circ} = 30^{\circ}$ (adj. ∠s on a str. line).

$n = \frac{360^{\circ}}{30^{\circ}} = 12$ (ext. ∠ of regular polygon)

Check: $90^{\circ} + 120^{\circ} + 150^{\circ} = 360^{\circ}$ , and a regular polygon with 12 sides has each interior angle $\frac{(12 - 2) \times 180^{\circ}}{12} = 150^{\circ}$ .

The third tile has 12 sides.

Key Points

An exterior angle of a convex polygon is the angle between one side and the extension of an adjacent side, at the vertex where they meet; take one at each vertex, all the same way round.
At each vertex of a convex polygon, interior angle + exterior angle = 180° (adj. ∠s on a str. line).
The exterior angles of any convex polygon add up to 360° (sum of ext. ∠s of polygon), whatever the number of sides; the interior angles add up to $(n - 2) \times 180^{\circ}$ (∠ sum of polygon).
For a regular polygon with $n$ sides: each exterior angle $= \frac{360^{\circ}}{n}$ (ext. ∠ of regular polygon), and each interior angle $= 180^{\circ} -$ each exterior angle.
Number of sides $= 360^{\circ} \div$ each exterior angle. From an interior angle, find the exterior angle first.
An angle can be each exterior angle of a regular polygon exactly when $360^{\circ} \div$ that angle is a whole number, 3 or more; that whole number is the number of sides. To test an interior angle, test the exterior angle beside it. Each exterior angle is at most 120°, and each interior angle is at least 60° and less than 180°.
Regular pentagon: 108° and 72°; hexagon: 120° and 60°; octagon: 135° and 45°; decagon: 144° and 36° (interior and exterior). A regular polygon with $n$ sides has $n$ lines of symmetry and rotational symmetry of order $n$ .
Where shapes meet at a point, their angles add up to 360° (∠s at a pt): three regular hexagons fit ( $3 \times 120^{\circ}$ ), but regular pentagons do not, since 360° ÷ 108° is not a whole number.