Angles on a line and around a point, angles in parallel lines, angles in triangles and quadrilaterals, and interior and exterior angles of polygons.

What this dot point is asking

AQA wants you to apply the basic angle rules for lines and points, the parallel-line angle relationships, the angle sums in triangles and quadrilaterals, and the interior and exterior angle rules for polygons, often chaining several together in a single multi-step problem. Crucially, AQA wants reasons: many marks are awarded for stating the named rule, not just the number.

Angles on a line and around a point

Angles that meet on a straight line add up to $180^\circ$, so if one is $130^\circ$ the other is $50^\circ$. Angles around a point add up to $360^\circ$. Vertically opposite angles, formed where two lines cross, are equal. These three rules unlock most basic angle-chasing.

Angles in parallel lines

When a straight line (a transversal) crosses two parallel lines, three relationships appear:

• Corresponding angles (in matching positions, an F-shape) are equal.
• Alternate angles (on opposite sides of the transversal, a Z-shape) are equal.
• Co-interior angles (between the parallel lines on the same side, a C-shape) sum to $180^\circ$.

AQA expects the correct name as the reason. Writing "alternate angles are equal" earns the reasoning mark; writing only the number does not.

Angles in triangles and quadrilaterals

The interior angles of a triangle sum to $180^\circ$, and an exterior angle of a triangle equals the sum of the two opposite interior angles. An isosceles triangle has two equal base angles. A quadrilateral has interior angles summing to $360^\circ$.

Interior and exterior angles of polygons

For a regular hexagon ($n = 6$), each exterior angle is $\dfrac{360^\circ}{6} = 60^\circ$ and each interior angle is $120^\circ$. To go the other way, if a regular polygon has an exterior angle of $36^\circ$, then $n = \dfrac{360^\circ}{36^\circ} = 10$ sides.

The interior-angle sum from triangles

It is worth seeing why the interior-angle sum is $(n - 2) \times 180^\circ$. Any polygon can be split into triangles by drawing diagonals from one vertex: a quadrilateral splits into two triangles, a pentagon into three, and in general an $n$-sided polygon into $(n - 2)$ triangles. Since each triangle contributes $180^\circ$, the total is $(n - 2) \times 180^\circ$. So a pentagon has interior angles summing to $3 \times 180^\circ = 540^\circ$, and a regular pentagon has each angle $\dfrac{540^\circ}{5} = 108^\circ$. Understanding the derivation means you never have to memorise the formula blindly.

Tessellation and angle reasoning

Whether regular polygons tessellate (tile a flat surface with no gaps) comes straight from the interior angle. Shapes tessellate when copies of their interior angles fit exactly around a point, summing to $360^\circ$. Equilateral triangles ($60^\circ$), squares ($90^\circ$) and regular hexagons ($120^\circ$) all divide into $360^\circ$ a whole number of times, so they tessellate, but a regular pentagon ($108^\circ$) does not, since $108$ does not divide $360$ exactly. This is a neat application that ties the polygon angle rules to a visual result examiners like to ask about.