// maths › Non-Right-Angle Trigonometry
Derive a triangle's area through trigonometry: use the cosine rule to find the included angle, then the area formula ½·a·b·sin C, and watch it simplify to Heron's formula — showing why the two routes give the same area.
Area = ½·a·b·sin C, with cos C = (a²+b²−c²)/(2ab) → Heron's √(s(s−a)(s−b)(s−c))
The geometry version applies the finished formula √(s(s−a)(s−b)(s−c)) directly. This trigonometry version derives the area instead: it uses the cosine rule to find the included angle, then the area formula ½·a·b·sin C, and shows that the result is exactly Heron's formula.
Substitute cos C = (a²+b²−c²)/(2ab) from the cosine rule into sin C = √(1−cos²C), then into ½·a·b·sin C. After simplifying, the expression collapses to √(s(s−a)(s−b)(s−c)) — Heron's formula. The two are algebraically identical, so they always agree.
For a 3-4-5 triangle, the angle between sides 3 and 4 has cos C = (9+16−25)/(2·3·4) = 0, so C = 90° and sin C = 1. Then Area = ½·3·4·1 = 6 — the same as Heron's √(6·3·2·1) = 6 square units.
C is the angle between sides a and b, the side opposite it being c. The cosine rule pins it down from the three side lengths. It is only an intermediate step — what the area formula actually needs is sin C.
The cosine-rule-to-area route is common in surveying and navigation, where two distances and the angle between them are known, and in computer graphics and physics, where triangle areas are computed from vectors whose included angle comes from a dot product.