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Cambridge 0580 IGCSE  ·  Topic C4.6

C4.6 Angles

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01 — Explanation

What basic angle facts do I need?

Every angle-chasing problem is built from a handful of core facts, and each one must be stated as a reason to earn full marks:

  • Angles at a point sum to 360∘360^\circ.
  • Angles on a straight line sum to 180∘180^\circ.
  • Vertically opposite angles are equal.
  • Angles in a triangle sum to 180∘180^\circ.
  • Angles in a quadrilateral sum to 360∘360^\circ.

Exam tip

Cambridge awards separate reason marks. A correct numeric angle with no justification loses marks — always write the property used, e.g. "co-interior angles sum to 180∘180^\circ," and use three-letter notation (angle ABCABC) when points are labelled.

02 — Explanation

How do I spot angles formed by parallel lines?

When a transversal crosses two parallel lines, three relationships appear:

        ╱              Alternate angles (Z-shape):     equal
  ─────╱───── Line 1    Corresponding angles (F-shape): equal
      ╱                 Co-interior angles (C-shape):   sum to 180°
  ───╱────── Line 2
    ╱
RelationshipShape formedRule
Alternate anglesZ-shapeEqual
Corresponding anglesF-shapeEqual
Co-interior anglesC-shapeSum to 180∘180^\circ

Tracing the letter shape with your finger over the diagram is a reliable way to identify which rule applies under time pressure — if it looks like a "Z", the angles are equal; if it looks like a "C" or "U", they sum to 180∘180^\circ.

03 — Explanation

How do I find the interior and exterior angle of a regular polygon?

For any polygon with nn sides:

sum of interior angles=(n−2)×180∘sum of exterior angles=360∘ (always, for any polygon)\text{sum of interior angles} = (n-2)\times 180^\circ \qquad \text{sum of exterior angles} = 360^\circ \ \text{(always, for any polygon)}

The (n−2)×180∘(n-2)\times180^\circ rule comes from splitting the polygon into triangles by drawing diagonals from one vertex: an nn-sided polygon always splits into exactly n−2n-2 triangles this way, and each triangle contributes 180∘180^\circ to the total. The exterior-angle sum of 360∘360^\circ holds for any polygon, regular or not, because walking all the way around the outside and turning through each exterior angle in turn brings you back to face your starting direction — one full turn, every time.

For a regular polygon, divide by nn to get a single angle:

one exterior angle=360∘none interior angle=180∘−360∘n\text{one exterior angle} = \frac{360^\circ}{n} \qquad \text{one interior angle} = 180^\circ - \frac{360^\circ}{n}

This division by nn only works when the polygon is regular — for an irregular polygon, only the angle sum is fixed, not the individual angles, since the sides and angles aren't all equal.

Worked example. A regular polygon has an exterior angle of 24∘24^\circ. Find (a) its number of sides, (b) its interior angle.

(a) n=360∘÷24∘=15n = 360^\circ \div 24^\circ = 15 sides.

(b) Interior and exterior angles lie on a straight line: 180∘−24∘=156∘180^\circ - 24^\circ = 156^\circ.

Worked example. In the diagram, two parallel lines are crossed by a transversal, creating a co-interior angle of x+30∘x + 30^\circ and its partner of 2x2x. Find xx.

Co-interior angles sum to 180∘180^\circ:

(x+30)+2x=1803x+30=1803x=150x=50∘(x+30) + 2x = 180 \qquad 3x + 30 = 180 \qquad 3x = 150 \qquad x = 50^\circ

Common mistake

assuming every pair of angles formed by a transversal is either equal or supplementary without checking which shape (Z, F, or C) it actually forms. Always identify the relationship first, then apply the matching rule.