An exterior angle of a triangle is formed by extending one side of the triangle beyond a vertex. The exterior angle theorem states that this angle equals the sum of the two interior angles that are not adjacent to it — the two non-adjacent (or remote) interior angles. This result follows directly from the angle-sum rule for triangles.
What is an exterior angle of a triangle?
An interior angle of a triangle is the angle inside the triangle at each vertex. An exterior angle is formed at a vertex by extending one side of the triangle outward in a straight line. At each vertex there are two possible exterior angles (one on each side of the extended line), but they are equal — they are vertically opposite.
At a given vertex, the interior angle and one exterior angle together form a straight line, so they add up to 180°.
For example, if a triangle has interior angles of 50°, 70°, and 60°:
- At the vertex with 60°, the exterior angle = 180° − 60° = 120°.
- Check using the theorem: 50° + 70° = 120°. ✓
What does the exterior angle theorem state?
Exterior angle theorem: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
In notation: if a triangle has interior angles A, B, and C, and an exterior angle e is formed at vertex C, then:
e = A + B
This works at any vertex. The "non-adjacent" angles are the two interior angles at the other two vertices — the ones not touching the extended side.
How do you prove the exterior angle theorem?
Proof using angles on a straight line and angle sum of a triangle:
- The interior angles of any triangle add to 180°: A + B + C = 180°.
- The exterior angle e and the adjacent interior angle C lie on a straight line: e + C = 180°.
- From (1): A + B = 180° − C.
- Substituting into (2): e = 180° − C = A + B. ∎
This is a simple two-step proof that is worth remembering at KS3, because examiners may ask you to "give a reason" for your working.
How do you use the exterior angle theorem to find missing angles?
Worked example 1: In triangle PQR, angle P = 48° and angle Q = 65°. Side QR is extended beyond R to a point S. Find angle QRS (the exterior angle at R).
Using the theorem: angle QRS = angle P + angle Q = 48° + 65° = 113°
Check: angle PRQ = 180° − 48° − 65° = 67°. Exterior angle = 180° − 67° = 113°. ✓
Worked example 2: An exterior angle of a triangle is 95°. One of the non-adjacent interior angles is 40°. Find the other non-adjacent interior angle.
Let the unknown angle = x. 95° = 40° + x → x = 95° − 40° = 55°
Check: third interior angle = 180° − 55° − 40° = 85°. Exterior angle = 180° − 85° = 95°. ✓
Worked example 3 (algebra): An exterior angle is (3x + 10)°. The two non-adjacent interior angles are 2x° and (x + 20)°.
(3x + 10) = 2x + (x + 20) 3x + 10 = 3x + 20
This gives 10 = 20, which is a contradiction. That means the angles as written cannot all exist simultaneously in one triangle — check the original problem for a typo.
A corrected version: exterior angle = (4x + 10)°, non-adjacent angles = 2x° and (x + 20)°. 4x + 10 = 2x + x + 20 → 4x + 10 = 3x + 20 → x = 10. Angles: 20°, 30°, exterior = 50°. Third interior = 180° − 20° − 30° = 130°. Exterior = 180° − 130° = 50°. ✓
What is the sum of the exterior angles of a triangle?
If you form one exterior angle at each of the three vertices, the three exterior angles add to 360°.
This is true for any convex polygon: the sum of one exterior angle per vertex is always 360°. For a triangle specifically, each exterior angle + its adjacent interior angle = 180°, so the sum of the three exterior angles = 3 × 180° − (sum of interior angles) = 540° − 180° = 360°.
| Shape | Sum of interior angles | Sum of one exterior angle per vertex |
|---|---|---|
| Triangle | 180° | 360° |
| Quadrilateral | 360° | 360° |
| Pentagon | 540° | 360° |
| Any convex polygon | (n−2) × 180° | 360° |
Frequently asked questions
Do I need to use the theorem, or can I just use the angle sum of a triangle instead?
Both methods give the same answer, but the theorem is quicker when you are given two interior angles and asked for the exterior angle — you just add them, rather than finding the third interior angle first then subtracting from 180°. In exam papers, a question that asks you to "use the exterior angle theorem" or to "give a reason" expects you to name the theorem explicitly.
Can the exterior angle theorem be used for obtuse triangles?
Yes. The theorem applies to all triangles regardless of whether the angles are acute, right, or obtuse. The proof used only the angle sum of a triangle (always 180°) and angles on a straight line (always 180°), both of which hold for any triangle.
How is the exterior angle theorem different from angles in polygons?
The exterior angle theorem applies specifically to the relationship between one exterior angle and the two non-adjacent interior angles of a triangle. The polygon rule (sum of exterior angles = 360°) applies more broadly but does not tell you about individual interior angle relationships. They are related but distinct results.
What reason should I write in an exam?
Write: "Exterior angle of a triangle equals the sum of the two non-adjacent interior angles." You may abbreviate this to "exterior angle of a triangle" once you have stated the theorem, but write it in full the first time.
For Socratic KS3 geometry practice on triangle angles with Professor Pi, see aitutors.me.