Angle Rules and Polygons

1 The Basic Angle Facts

Learn these four

Angles on a straight line add to $180^\circ$
Angles at a point add to $360^\circ$
Vertically opposite angles are equal
Angles in a triangle add to $180^\circ$

Worked Example 1 — Angles at a point

Four angles meet at a point. Three of them are $85^\circ$, $112^\circ$ and $63^\circ$. Find the fourth.

①Angles at a point add to $360^\circ$.

②$85 + 112 + 63 = 260$

③Fourth angle $= 360 - 260 = 100^\circ$

Worked Example 2 — Using algebra

Angles on a straight line are $(3x + 10)^\circ$, $(2x)^\circ$ and $(x + 20)^\circ$. Find $x$ and hence the largest angle.

①$(3x + 10) + 2x + (x + 20) = 180$

②$6x + 30 = 180$

③$6x = 150 \Rightarrow x = 25$

④Angles: $3(25)+10 = 85^\circ$, $2(25) = 50^\circ$, $25 + 20 = 45^\circ$.

Largest angle $= 85^\circ$. Check: $85 + 50 + 45 = 180$ ✓

2 Angles in Parallel Lines

When a straight line (a transversal) crosses a pair of parallel lines, three special relationships appear.

Name	Shape to look for	Rule
Alternate	Z	Equal
Corresponding	F	Equal
Co-interior (allied)	C or U	Add to $180^\circ$

In the diagram above:
• Corresponding (F-shape): $a = e$, $b = f$, $c = g$, $d = h$
• Alternate (Z-shape): $c = e$ and $d = f$ (both between the parallel lines, on opposite sides)
• Co-interior (C-shape): $c + f = 180^\circ$ and $d + e = 180^\circ$

The lines must actually be parallel — look for the matching arrows. Without them, none of these rules applies.

Worked Example 3 — A parallel-line chase

Two parallel lines are crossed by a transversal. One angle is $118^\circ$. Find the co-interior angle and the corresponding angle.

①Corresponding angles are equal, so the corresponding angle is $118^\circ$.

②Co-interior angles add to $180^\circ$.

③Co-interior angle $= 180 - 118 = 62^\circ$.

If you get stuck on a parallel-line problem, draw an extra line through the awkward point parallel to the given pair. It turns one hard angle into two easy alternate angles.

3 Angles in Triangles

Two essential facts

Angle sum of a triangle $= 180^\circ$
Exterior angle $=$ sum of the two opposite interior angles

Why the exterior angle rule works. $b + e = 180^\circ$ (angles on a straight line) and $a + b + c = 180^\circ$ (angle sum). Setting these equal gives $b + e = a + b + c$, so $e = a + c$.

Worked Example 4 — Isosceles triangle

An isosceles triangle has an apex angle of $34^\circ$. Find the base angles.

①The two base angles are equal. Call each $x$.

②$34 + x + x = 180$

③$2x = 146 \Rightarrow x = 73^\circ$

Two possible answers. If a question says "an isosceles triangle has an angle of $40^\circ$", that $40^\circ$ could be the apex (giving base angles $70^\circ$, $70^\circ$) or a base angle (giving $40^\circ$, $40^\circ$, $100^\circ$). Check whether the question tells you which.

4 Interior Angles of a Polygon

Interior angle sum

$\text{sum} = (n - 2) \times 180^\circ$ for a polygon with $n$ sides

Why? Pick one vertex and draw every diagonal from it. This splits an $n$-sided polygon into exactly $n - 2$ triangles, and each triangle contributes $180^\circ$.

Sides $n$	Name	Interior sum	Each interior angle if regular
$3$	Triangle	$180^\circ$	$60^\circ$
$4$	Quadrilateral	$360^\circ$	$90^\circ$
$5$	Pentagon	$540^\circ$	$108^\circ$
$6$	Hexagon	$720^\circ$	$120^\circ$
$8$	Octagon	$1080^\circ$	$135^\circ$
$10$	Decagon	$1440^\circ$	$144^\circ$
$12$	Dodecagon	$1800^\circ$	$150^\circ$

5 Exterior Angles of a Polygon

The most useful polygon fact

The exterior angles of any polygon add to $360^\circ$
For a regular $n$-gon: each exterior angle $= \dfrac{360^\circ}{n}$

Why $360^\circ$? Imagine walking all the way round the outside of the polygon. At each corner you turn through the exterior angle. When you arrive back at the start facing the original direction, you have turned through exactly one full circle.

The link between the two

interior angle $+$ exterior angle $= 180^\circ$ (they lie on a straight line)

Nearly always use the exterior angle first. Finding $360 \div n$ and subtracting from $180$ is quicker and safer than $(n-2)\times 180 \div n$.

Worked Example 5 — Finding an interior angle

Find each interior angle of a regular polygon with $15$ sides.

①Exterior angle $= 360 \div 15 = 24^\circ$

②Interior angle $= 180 - 24 = 156^\circ$

Check: $(15-2)\times 180 = 2340^\circ$; $2340 \div 15 = 156^\circ$ ✓

Worked Example 6 — Working backwards to find $n$

A regular polygon has interior angles of $162^\circ$. How many sides does it have?

①Exterior angle $= 180 - 162 = 18^\circ$

②$n = 360 \div 18 = 20$

The polygon has $20$ sides.

Worked Example 7 — An irregular polygon

A hexagon has five interior angles of $110^\circ$, $135^\circ$, $98^\circ$, $142^\circ$ and $121^\circ$. Find the sixth.

①Interior sum $= (6-2) \times 180 = 720^\circ$

②$110 + 135 + 98 + 142 + 121 = 606$

③Sixth angle $= 720 - 606 = 114^\circ$

6 Why Only Some Shapes Tessellate

Shapes tessellate when they fit together with no gaps and no overlaps. For regular polygons this happens only when the interior angle divides exactly into $360^\circ$.

Regular shape	Interior angle	$360 \div$ angle	Tessellates?
Triangle	$60^\circ$	$6$	Yes
Square	$90^\circ$	$4$	Yes
Pentagon	$108^\circ$	$3.33\ldots$	No
Hexagon	$120^\circ$	$3$	Yes
Octagon	$135^\circ$	$2.66\ldots$	No

Only the equilateral triangle, the square and the regular hexagon tessellate on their own. That is why honeycombs and floor tiles use those shapes.

7 Quick Reference

Straight line

Angles add to $180^\circ$.

At a point

Angles add to $360^\circ$.

Vertically opposite

Equal — the X-shape.

Alternate (Z)

Equal.

Corresponding (F)

Equal.

Co-interior (C)

Add to $180^\circ$.

Triangle

Sum $180^\circ$; exterior $=$ sum of the two opposite interior angles.

Polygon interior

Sum $= (n-2)\times 180^\circ$.

Polygon exterior

Sum $= 360^\circ$; each one is $\dfrac{360}{n}$ if regular.

Always

Write the reason next to every angle you find.

8 Practice Questions

Question 1

Angles at a point are $72^\circ$, $x$, $118^\circ$ and $2x$. Find $x$.

▶ Show solution

Angles at a point add to $360^\circ$.

$72 + x + 118 + 2x = 360$

$3x + 190 = 360 \Rightarrow 3x = 170$

$x = 56.67^\circ$ (2 d.p.), or exactly $\dfrac{170}{3}^\circ$.

Question 2

Two parallel lines are crossed by a transversal. One of the angles is $73^\circ$. Write down, with reasons, the sizes of the corresponding, alternate and co-interior angles.

▶ Show solution

Corresponding angle $= 73^\circ$ — corresponding angles are equal.

Alternate angle $= 73^\circ$ — alternate angles are equal.

Co-interior angle $= 180 - 73 = 107^\circ$ — co-interior angles add to $180^\circ$.

Question 3

Find the interior angle sum of (a) a nonagon (9 sides), (b) a 20-sided polygon.

▶ Show solution

(a) $(9-2)\times 180 = 7 \times 180 = 1260^\circ$

(b) $(20-2)\times 180 = 18 \times 180 = 3240^\circ$

Question 4

A regular polygon has exterior angles of $20^\circ$. Find (a) the number of sides, (b) each interior angle, (c) the interior angle sum.

▶ Show solution

(a) $n = 360 \div 20 = 18$ sides.

(b) Interior $= 180 - 20 = 160^\circ$.

(c) Sum $= (18-2)\times 180 = 2880^\circ$. Check: $18 \times 160 = 2880$ ✓

Question 5

In triangle $ABC$, side $BC$ is extended to $D$. Angle $ABC = 52^\circ$ and angle $BAC = 61^\circ$. Find angle $ACD$ using the exterior angle rule, and check your answer another way.

▶ Show solution

Exterior angle $ACD$ = sum of the two opposite interior angles.

$\angle ACD = 52 + 61 = 113^\circ$

Check: $\angle ACB = 180 - 52 - 61 = 67^\circ$; angles on a straight line give $180 - 67 = 113^\circ$ ✓

Question 6

A regular polygon has interior angles of $140^\circ$. How many sides has it?

▶ Show solution

Exterior angle $= 180 - 140 = 40^\circ$

$n = 360 \div 40 = 9$ sides (a nonagon).

Question 7

An isosceles triangle has one angle of $110^\circ$. Find the other two angles, and explain why there is only one possible answer.

▶ Show solution

$110^\circ$ cannot be one of the equal pair, because two angles of $110^\circ$ would total $220^\circ$ — already more than $180^\circ$.

So $110^\circ$ is the apex angle.

Base angles: $(180 - 110) \div 2 = 70 \div 2 = 35^\circ$ each.

Angles: $110^\circ$, $35^\circ$, $35^\circ$.

Question 8

The interior angles of a pentagon are $x$, $2x$, $2x$, $3x$ and $4x$. Find the size of the largest angle.

▶ Show solution

Interior sum of a pentagon $= (5-2)\times 180 = 540^\circ$.

$x + 2x + 2x + 3x + 4x = 540$

$12x = 540 \Rightarrow x = 45$

Largest angle $= 4x = 180^\circ$.

Note: an interior angle of exactly $180^\circ$ means that "vertex" is actually a straight line, so this shape is really a quadrilateral — a good reminder to sense-check an answer.

Question 9

Explain, using angle facts, why regular pentagons cannot tessellate but regular hexagons can.

▶ Show solution

Shapes meeting at a point must have interior angles adding to exactly $360^\circ$.

Pentagon: interior angle $= 180 - (360 \div 5) = 180 - 72 = 108^\circ$. $360 \div 108 = 3.33\ldots$, not a whole number, so three pentagons leave a gap of $360 - 324 = 36^\circ$ and four would overlap.

Hexagon: interior angle $= 180 - 60 = 120^\circ$. $360 \div 120 = 3$ exactly, so three hexagons meet perfectly at every point.

Question 10

A regular hexagon and a regular polygon with $n$ sides meet at a point, together with an equilateral triangle, with no gaps and no overlaps.

(a) Write an equation for the angles at that point. (b) Find $n$ and name the polygon.

▶ Show solution

(a) Hexagon interior angle $= 120^\circ$; equilateral triangle $= 60^\circ$.

Let the third polygon's interior angle be $I$. Then:

$120 + 60 + I = 360$

(b) $I = 360 - 180 = 180^\circ$.

An interior angle of $180^\circ$ is impossible for a polygon, so no such polygon exists — the hexagon and triangle alone already leave a straight-line gap.

A valid alternative: two hexagons and one triangle give $120 + 120 + 60 = 300^\circ$, which also fails. But one hexagon and four triangles work: $120 + 4(60) = 360^\circ$ ✓ — a genuine semi-regular tessellation.

Angle Rules & Polygons (G3) · GCSE Maths Revision · Created with MathJax