Angle Sum of Polygon
Diagonals from one vertex split a polygon with \(n\) sides into \(n - 2\) triangles, so its interior angles add to \(180(n - 2)^\circ\). This NSW Year 9 Advanced maths subtopic uses that rule for any polygon (NESA outcome MA5‑GEO‑P‑01).
You will find angle sums and the angles of regular polygons, find the number of sides from an angle sum or from one interior angle, and find unknown angles in pentagons, hexagons and an arrowhead. You will decide whether an angle sum of \(1000^\circ\) is possible, find the angles where a square and a pentagon or three tiles meet, and explain the rule using a heptagon. It is part of Year 9 Advanced Properties of Geometric Figures.
Every question with a fully worked solution.
- Angle Sum of Polygon - Video - Interior Angles of a Polygon Watch
Theory
The angle sum of a polygon with \(n\) sides is \(180(n - 2)^\circ\), because the diagonals from one vertex cut it into \(n - 2\) triangles. This Year 9 Advanced maths page uses the formula forwards and backwards: interior angles of regular polygons, the number of sides from an angle, reflex angles, angles written as expressions, and shapes that meet at a point.
A polygon is a closed shape made of straight sides. Its interior angles are the angles inside it at each vertex. A polygon is convex if every interior angle is less than \(180^\circ\), and concave if one or more is a reflex angle (between \(180^\circ\) and \(360^\circ\)).
From one vertex of a convex polygon with \(n\) sides you can draw \(n - 3\) diagonals. They cut it into \(n - 2\) triangles whose angles exactly fill the polygon's angles. So the angle sum is \(180(n - 2)^\circ\).
A regular polygon has all sides equal and all angles equal. Each interior angle is the angle sum divided by \(n\).
The exterior angle at a vertex is the angle between one side and the next side extended. It makes a straight line with the interior angle, so interior \(+\) exterior \(= 180^\circ\). The exterior angles of a convex polygon add to \(360^\circ\), which gives a fast route to the number of sides of a regular polygon.
Where shapes meet at a point with no gaps or overlaps, their angles add to \(360^\circ\).
For a polygon with \(n\) sides:
For a regular polygon with \(n\) sides, each interior angle and each exterior angle are:
| Sides \(n\) | Name | Triangles | Angle sum |
|---|---|---|---|
| \(3\) | triangle | \(1\) | \(180^\circ\) |
| \(4\) | quadrilateral | \(2\) | \(360^\circ\) |
| \(5\) | pentagon | \(3\) | \(540^\circ\) |
| \(6\) | hexagon | \(4\) | \(720^\circ\) |
| \(n\) | \(n\)-gon | \(n - 2\) | \(180(n - 2)^\circ\) |
How to solve a polygon angle problem
- Count the sides \(n\) and work out the angle sum \(180(n - 2)\).
- Write every interior angle, using a reflex angle where the shape is pushed in, and set their total equal to the angle sum.
- Solve the equation, then substitute back to find the angle the question asks for.
- Check that the angles add to the angle sum.
Regular polygons
- Angle to sides: exterior angle \(= 180^\circ -\) interior angle, then \(n = \dfrac{360}{\text{exterior}}\).
- Sides to angle: interior angle \(= \dfrac{180(n - 2)}{n}\), or \(180^\circ - \dfrac{360^\circ}{n}\).
- One angle different: if a polygon has one angle of \(a^\circ\) and \(n - 1\) angles of \(b^\circ\), solve \(a + b(n - 1) = 180(n - 2)\).
- Two regular polygons linked: write each interior (or exterior) angle in terms of \(n\) and form one equation. Exterior angles usually give simpler fractions.
- Diagonals: two equal sides of a regular polygon and a diagonal make an isosceles triangle, so its base angles are \(\dfrac{180^\circ - \text{interior}}{2}\).
- Meeting at a point: the angles of shapes that fit round a point add to \(360^\circ\).
(i) A polygon has an angle sum of \(1980^\circ\). How many sides does it have?
(ii) Can the angles of a polygon add to \(2000^\circ\)? Explain.
(i) Put the angle sum into \(S = 180(n - 2)\) and solve for \(n\).
| \(180(n - 2)\) | \(=\) | \(1980\) |
| \(n - 2\) | \(=\) | \(\dfrac{1980}{180} = 11\) |
| \(n\) | \(=\) | \(\mathbf{13}\) |
(ii) Try the same steps with \(2000^\circ\).
| \(n - 2\) | \(=\) | \(\dfrac{2000}{180} = 11.11\ldots\) |
| \(n\) | \(=\) | \(13.11\ldots\) |
A polygon must have a whole number of sides, so no. Every angle sum is a multiple of \(180^\circ\): \(13\) sides give \(1980^\circ\) and \(14\) sides give \(2160^\circ\), and \(2000^\circ\) lies between them.
Method 1: interior angle formula.
| \(\dfrac{180(n - 2)}{n}\) | \(=\) | \(156\) |
| \(180n - 360\) | \(=\) | \(156n\) |
| \(24n\) | \(=\) | \(360\) |
| \(n\) | \(=\) | \(\mathbf{15}\) |
Method 2: exterior angle. The interior and exterior angles lie on a straight line, and the exterior angles of a polygon add to \(360^\circ\).
| \(e\) | \(=\) | \(180^\circ - 156^\circ = 24^\circ\) |
| \(n\) | \(=\) | \(\dfrac{360}{24} = \mathbf{15}\) |
Method 2 is shorter and avoids the equation.
The angle sum of a pentagon is \(180 \times 3 = 540^\circ\). It still applies when the pentagon is concave, as long as the reflex angle is used.
| \(90 + 90 + x + x + (2x + 60)\) | \(=\) | \(540\) |
| \(4x + 240\) | \(=\) | \(540\) |
| \(4x\) | \(=\) | \(300\) |
| \(x\) | \(=\) | \(\mathbf{75}\) |
| \(\text{reflex } \angle D\) | \(=\) | \(2(75) + 60 = \mathbf{210^\circ}\) |
Check: \(90 + 90 + 75 + 75 + 210 = 540\).
Each interior angle of a regular hexagon is \(\dfrac{180 \times 4}{6} = 120^\circ\). \(AB = BC\), so \(\triangle ABC\) is isosceles.
| \(\angle BAC\) | \(=\) | \(\dfrac{180^\circ - 120^\circ}{2}\) |
| \(\) | \(=\) | \(\mathbf{30^\circ}\) |
(angles opposite equal sides, angle sum of a triangle)
| \(\angle FAC\) | \(=\) | \(\angle FAB - \angle BAC\) |
| \(\) | \(=\) | \(120^\circ - 30^\circ\) |
| \(\) | \(=\) | \(90^\circ\) |
So \(AC \perp AF\).
Common pitfalls
Frequently asked questions
What is the formula for the angle sum of a polygon?
The angle sum of a polygon with \(n\) sides is \(180(n - 2)^\circ\). A pentagon has \(180 \times 3 = 540^\circ\).
Why is the angle sum 180(n - 2)?
The diagonals from one vertex split the polygon into \(n - 2\) triangles. Each triangle has an angle sum of \(180^\circ\), and together their angles make up exactly the angles of the polygon.
How do you find the number of sides of a regular polygon from one angle?
Subtract the interior angle from \(180^\circ\) to get the exterior angle, then divide \(360\) by the exterior angle. An interior angle of \(156^\circ\) gives an exterior angle of \(24^\circ\) and \(15\) sides.
Does the angle sum formula work for concave polygons?
Yes. A concave polygon with \(n\) sides still has an angle sum of \(180(n - 2)^\circ\), as long as you use the reflex angle inside the shape at the pushed-in vertex.
Can a polygon have any angle sum?
No. The angle sum is always a multiple of \(180^\circ\), starting at \(180^\circ\) for a triangle. If \(180(n - 2)\) cannot equal the given total for a whole number \(n\), no polygon has that angle sum.
Which regular polygons tile a floor on their own?
Only equilateral triangles, squares and regular hexagons. Their interior angles, \(60^\circ\), \(90^\circ\) and \(120^\circ\), divide exactly into \(360^\circ\), so copies fit round a point with no gap.