Triangle Geometry
The angles of a triangle add to \(180^\circ\), and an exterior angle equals the sum of the two interior opposite angles. This NSW Year 9 Advanced maths subtopic revises these Stage 4 properties (NESA outcome MA4‑GEO‑C‑01) and uses them with parallel lines and isosceles triangles.
You will find unknown angles in diagrams, giving reasons, and find angles from a ratio such as \(2 : 3 : 5\). Algebra questions set up equations for angles like \((2x + 15)^\circ\). You will find the angle between a roof brace and a rafter and the angle where two bisectors meet, prove the exterior angle result with a parallel line, and solve a puzzle about isosceles triangles. It is part of Year 9 Advanced Properties of Geometric Figures.
Theory
The angles of a triangle add to 180°, and an exterior angle equals the sum of the two interior opposite angles. This Year 9 Advanced maths page revises the triangle properties from earlier years and takes them further: isosceles triangles, parallel lines, angles written with algebra, bisectors and proofs with reasons.
Triangles are classified by their sides: scalene (no equal sides), isosceles (two equal sides) and equilateral (three equal sides). They are also classified by their angles: acute-angled (all angles less than \(90^\circ\)), right-angled (one \(90^\circ\) angle) and obtuse-angled (one angle more than \(90^\circ\)). A triangle has one name of each kind, such as a right-angled isosceles triangle.
\(\triangle ABC\) names the triangle by its vertices. \(\angle ABC\) is the angle at the middle letter, \(B\).
The angle sum of a triangle is \(180^\circ\). In a right-angled triangle the two acute angles add to \(90^\circ\).
If one side is extended, the angle outside the triangle is an exterior angle. It equals the sum of the two interior opposite angles (the two not next to it).
In an isosceles triangle the angles opposite the equal sides (the base angles) are equal. Every angle of an equilateral triangle is \(60^\circ\).
The facts about angles at a point, on a straight line, vertically opposite, and alternate, corresponding and co-interior angles on parallel lines are used alongside these. In a proof, every step needs a reason.
| Fact | Statement | Reason to write |
|---|---|---|
| Angle sum | \(\alpha + \beta + \gamma = 180^\circ\) | angle sum of \(\triangle\) |
| Exterior angle | exterior \(= \alpha + \beta\) | exterior angle of \(\triangle\) |
| Isosceles | base angles equal | base angles of isosceles \(\triangle\) |
| Straight line | adjacent angles add to \(180^\circ\) | angles on a straight line |
| Vertically opposite | equal | vertically opposite angles |
| Parallel lines | alternate and corresponding equal; co-interior add to \(180^\circ\) | alternate / corresponding / co-interior angles, \(AB \parallel CD\) |
How to find an unknown angle
- Mark the diagram: ticks for equal sides, arrows for parallel lines, every given angle.
- Look for the quickest fact: an exterior angle often saves a step over the angle sum.
- Fill in angles one at a time, writing a reason for each.
- Chain triangles: an angle found in one triangle is often a base angle or exterior angle of the next (Examples 2 and 3).
- Check that each triangle's angles add to \(180^\circ\).
Algebra and proof
- Angles with pronumerals: write an equation from one fact (angle sum, exterior angle or equal base angles), solve it, then substitute to find each angle (Example 1).
- No numbers given: let one angle be \(x\) and write the others in terms of it, such as an apex twice each base angle.
- Bisectors: a bisected angle splits into two equal halves. If only the sum of two angles is known, halve the sum.
- Proofs: draw a helpful extra line (often a parallel), state each equal pair with its reason, and finish with the result you were asked to prove (Example 4).
The exterior angle equals the sum of the two interior opposite angles.
| \(5x - 8\) | \(=\) | \((2x + 10) + (x + 22)\) |
| \(5x - 8\) | \(=\) | \(3x + 32\) |
| \(2x\) | \(=\) | \(40\) |
| \(x\) | \(=\) | \(\mathbf{20}\) |
So \(\angle KMN = 5(20) - 8 = 92^\circ\). \(\angle KML\) and \(\angle KMN\) are on a straight line:
| \(\angle KML\) | \(=\) | \(180^\circ - 92^\circ = \mathbf{88^\circ}\) |
Check with the angle sum: \(50^\circ + 42^\circ + 88^\circ\) \(= 180^\circ\).
| \(\angle PQR = \angle PRQ\) | \(=\) | \(\dfrac{180^\circ - 40^\circ}{2} = 70^\circ\) |
(base angles of isosceles \(\triangle PQR\), \(PQ = PR\))
| \(\angle QSR\) | \(=\) | \(\angle QRS = 70^\circ\) |
(base angles of isosceles \(\triangle QSR\), \(QS = QR\))
| \(\angle SQR\) | \(=\) | \(180^\circ - 70^\circ - 70^\circ = 40^\circ\) |
(angle sum of \(\triangle QSR\))
| \(\angle PQS\) | \(=\) | \(70^\circ - 40^\circ = \mathbf{30^\circ}\) |
Mark each pair of equal sides with ticks first; each pair gives a pair of equal angles.
| \(\angle BAC\) | \(=\) | \(180^\circ - 72^\circ - 46^\circ = 62^\circ\) |
(angle sum of \(\triangle ABC\))
| \(\angle BAD\) | \(=\) | \(\dfrac{62^\circ}{2} = 31^\circ\) |
(\(AD\) bisects \(\angle BAC\))
| \(\angle ADC\) | \(=\) | \(\angle ABD + \angle BAD\) |
| \(\) | \(=\) | \(72^\circ + 31^\circ = \mathbf{103^\circ}\) |
(exterior angle of \(\triangle ABD\))
Check in \(\triangle ADC\): \(31^\circ + 46^\circ + 103^\circ\) \(= 180^\circ\).
Draw a line \(DE\) through \(A\) parallel to \(BC\). Let the angles of the triangle be \(\alpha\), \(\beta\) and \(\gamma\).
| \(\angle DAB\) | \(=\) | \(\angle ABC = \beta\) |
(alternate angles, \(DE \parallel BC\))
| \(\angle EAC\) | \(=\) | \(\angle ACB = \gamma\) |
(alternate angles, \(DE \parallel BC\))
| \(\beta + \alpha + \gamma\) | \(=\) | \(180^\circ\) |
(angles on the straight line \(DAE\))
| \(\therefore\ \angle ABC + \angle BAC + \angle ACB\) | \(=\) | \(\mathbf{180^\circ}\) |
A proof works for every triangle because it uses letters and reasons, not measured angles.
Common pitfalls
Frequently asked questions
What do the angles in a triangle add up to?
The three interior angles of any triangle add to \(180^\circ\). In a right-angled triangle the two other angles add to \(90^\circ\).
What is the exterior angle theorem?
If one side of a triangle is extended, the exterior angle formed equals the sum of the two interior angles at the other two vertices.
How do you find the angles of an isosceles triangle?
The two base angles opposite the equal sides are equal. If you know the apex angle, subtract it from \(180^\circ\) and halve. If you know a base angle, the apex is \(180^\circ\) minus twice it.
Can a triangle be both right-angled and isosceles?
Yes. A right-angled isosceles triangle has a \(90^\circ\) angle and two \(45^\circ\) angles. A triangle has one side name and one angle name at the same time.
How do you prove the angle sum of a triangle?
Draw a line through one vertex parallel to the opposite side. The two alternate angles formed equal the other two angles of the triangle, and the three angles at the vertex make a straight angle of \(180^\circ\).