Triangle identities · derivation ·proof
Where A + B + C = 180° comes from
use it as an equation for the missing angle. In practice nobody proves this identity at the whiteboard; they solve it. Treat it as a one-step equation and watch the unit trap.
Building it step by step
- C = π − A − BRearrange for the unknown angle. · Angles of a triangle add to 180°
- A = π/3, B = π/4 → C = 5π/12π − π/3 − π/4 = 12π/12 − 4π/12 − 3π/12 = 5π/12 (75°).
- A = 100°, B = 90° → no triangleThe sum already exceeds 180° before C is chosen — the identity is also an existence test.
What this derivation depends on
Every line above is one of these — nothing else is assumed:
- A + B + C = 180° — Angles of a triangle add to 180°
That list is the argument for learning fewer formulas: most of this catalogue is a short chain away from the unit circle and the sum formulas.
Where this route is used
Two angles are known and the third must be found before any side work starts — this is the step that turns 'two angles' into a usable pair for the sine rule. It also tells you when a proposed triangle cannot exist.
- solve
- verify
Checked on these triangles
Deriving a formula and testing it are different things; this table is the test. The numbers come from the same engine as the verifier, computed when the site was built.
| Triangle (A) | Left side | Right side | Agree |
|---|---|---|---|
| 7.5° | 3.141593 | 3.141593 | yes |
| 15° | 3.141593 | 3.141593 | yes |
| 18° | 3.141593 | 3.141593 | yes |
| 22.5° | 3.141593 | 3.141593 | yes |
| 30° | 3.141593 | 3.141593 | yes |
| 37° | 3.141593 | 3.141593 | yes |
Neighbouring derivations
- a/sin A = b/sin B — Law of sines
- a² = b² + c² − 2bc·cos A — Law of cosines
- sin θ = cos(90° − θ) — Cofunction identity for sine
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from A + B + C = 180° and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the angles of a triangle add to 180° page.