• Angles at a point sum to $360°$.
• Angles on a straight line sum to $180°$.
• Vertically opposite angles (formed where two lines cross) are equal.
• Angles in a triangle sum to $180°$; angles in a quadrilateral sum to $360°$.

A line crossing a pair of parallel lines (a transversal) creates three named angle relationships:

• Corresponding angles (same position at each intersection, forming an "F" shape) are equal.
• Alternate angles (on opposite sides of the transversal, between the parallel lines, forming a "Z" shape) are equal.
• Co-interior angles (same side of the transversal, between the parallel lines, forming a "C" shape) are supplementary — they sum to $180°$.

The exterior angles of any polygon sum to $360°$, so for a regular polygon with $n$ sides:

Interior and exterior angles at each vertex lie on a straight line, so:

For an irregular polygon, use the sum-of-interior-angles formula instead:

Cambridge expects the correct geometric name for every step, using three-letter angle notation (e.g. angle $ABC$) where a diagram has multiple angles at one vertex. E.g.: "angle $ABC = 65°$ (alternate angles, $AB \parallel CD$)."

Work through the diagram one labelled angle at a time rather than trying to spot the final answer immediately. Use the parallel-line facts to find any angle you can from what's given, then feed that result into the triangle (or quadrilateral) angle-sum fact to find the next one, and so on — each new angle you calculate becomes a stepping stone for the next, so write down every intermediate value with its reason rather than trying to jump straight to the answer.

A regular polygon has an interior angle of $156°$. Find the number of sides.

1. Exterior angle $= 180° - 156° = 24°$.
2. Since exterior angles of a regular polygon sum to $360°$: $n = \dfrac{360°}{24°} = 15$.
3. The polygon has $15$ sides.

In a diagram, line $DE$ is parallel to line $BC$, and $A$ is a point above $DE$ such that $ABС$ forms a triangle with $D$ on $AB$ and $E$ on $AC$. Angle $ADE = 72°$ and angle $AED = 65°$. Find angle $DAE$, and hence state angle $ABC$.

1. Angles in triangle $ADE$ sum to $180°$: angle $DAE = 180° - 72° - 65° = 43°$.
2. Since $DE \parallel BC$, and $A$, $D$, $B$ lie on the same straight line (as do $A$, $E$, $C$), triangle $ABC$ is simply a larger version of triangle $ADE$ sharing the same angle at $A$.
3. Corresponding angles give angle $ABC =$ angle $ADE = 72°$ (since $AB$ is a transversal crossing the parallel lines $DE$ and $BC$).