Triangle identities · derivation ·proof
Where a/sin A = b/sin B comes from
same area, two different base/height pairings. Start from the area formula this site already publishes and cancel what is common — the ratio falls out without drawing a height at all.
Building it step by step
- ½·a·b·sin(C) = ½·b·c·sin(A)Both sides are the area of the same triangle. · Triangle area in two pairings
- a·sin(C) = c·sin(A)Divide both sides by ½b.
- a/sin(A) = c/sin(C)Divide by sin(A)·sin(C).
What this derivation depends on
Every line above is one of these — nothing else is assumed:
- ½ a b sin C = ½ b c sin A — Triangle area in two pairings
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
You have one complete pair — a side together with the angle opposite it — and you need an angle or a side from a second pair. It is also the formula that produces the ambiguous case: with two sides and a non-included angle, sin B can come from two different angles.
- solve
- evaluate
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° | 12.419529 | 12.419529 | yes |
| 15° | 3.006778 | 3.006778 | yes |
| 18° | 3.731834 | 3.731834 | yes |
| 22.5° | 5.219878 | 5.219878 | yes |
| 30° | 6.408182 | 6.408182 | yes |
| 37° | 4.162155 | 4.162155 | yes |
Neighbouring derivations
- a² = b² + c² − 2bc·cos A — Law of cosines
- ½ a b sin C = ½ b c sin A — Triangle area in two pairings
- A + B + C = 180° — Angles of a triangle add to 180°
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from ½ a b sin C = ½ b c sin A and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the law of sines page.