Triangle identities · proof ·what it is and when to use it
Proof: a = b·cos C + c·cos B
Projection rule — proved by split the side at the foot of the altitude. Every line below says which rule it uses, so nothing has to be taken on faith.
The proof
One altitude does it: side a is the sum of the two projections, and each projection is a right-triangle cosine read off the drawing.
- let D be the foot of the perpendicular from B onto ACThe altitude that splits side b — the same construction as in the sine rule.
- AD = c·cos A and DC = a·cos CRight-triangle cosine in ABD and in CBD.
- b = c·cos A + a·cos CAD + DC is side b.
- a = b·cos C + c·cos BWrite the same split for the altitude from C instead of the one from B. · Projection rule
Add the three versions and you get each side twice on both sides — which is why this rule is used to shorten proofs rather than to solve for unknowns.
Where the proof stops applying
The variables are one triangle's angles and sides, so they obey A + B + C = 180° and the sine rule. Drop that constraint and the equation is simply not true — which is why the sampler here generates triangles rather than random numbers.
The same claim, checked numerically
A proof is not the same thing as a check, and this page shows both: the reasoning above, and the first few triangles fed to the same engine behind the verifier. If they ever disagreed, the data would be broken — the build would fail before publishing (accuracy policy).
| Triangle (A) | Left side | Right side | Agree |
|---|---|---|---|
| 7.5° | 1.621074 | 1.621074 | yes |
| 15° | 0.778211 | 0.778211 | yes |
| 18° | 1.1532 | 1.1532 | yes |
| 22.5° | 1.997561 | 1.997561 | yes |
| 30° | 3.204091 | 3.204091 | yes |
| 37° | 2.504847 | 2.504847 | yes |
Related
- How would I find a = b·cos C + c·cos B myself? — the derivation, which is a different question from the proof.
- Projection rule: when to use it — the practical side.
- All triangle identities · proof index