Triangle Area with Sine

Whenever two sides and the INCLUDED angle are given, the classic SAS case. Then the formula delivers the area in one step, without you having to construct a height. Example: a = 8 cm, b = 6 cm, γ = 30°: A = ½·8·6·sin 30° = 24·0.5 = 12 cm². It is also practical when the height is hard to access, e.g. in surveying tasks with measured angles. If the included angle is missing, compute it first via the angle sum or the law of sines; if instead all three sides are given, Heron's formula is the more direct route. For γ = 90° the formula turns into the familiar A = ½·a·b.

Yes, necessarily. In A = ½·a·b·sin γ, γ is the angle ENCLOSED by a and b; only then is b·sin γ the height on a. If you insert one of the two other angles, you compute the area of a different triangle. Small numerical example: a triangle with a = 5, b = 7 and included γ = 60° has A = 17.5·sin 60° ≈ 15.16; wrongly taking an angle of 40° would give ≈ 11.25, clearly off. If only a non-included angle is known, first reduce it to the included one via the angle sum (α + β + γ = 180°) or the law of sines. A labelled sketch shows immediately which angle sits between the given sides.

It IS the same formula, just with the height computed. Choose a as the base g. The height h stands perpendicular on a and forms a right partial triangle with side b, in which h is the side opposite the angle γ: h = b·sin γ. Substituting into A = ½·g·h immediately gives A = ½·a·b·sin γ. So the sine takes over the height construction: instead of measuring or constructing h, you compute it from b and the angle. This also explains the limiting cases: at γ = 90°, sin γ = 1 and b itself is the height; for very small γ the triangle collapses, sin γ → 0 and the area vanishes.

With Heron's formula: A = √(s·(s − a)·(s − b)·(s − c)), where s = (a + b + c)/2 is the semi-perimeter. Example: sides 5, 6 and 7: s = 9, so A = √(9·4·3·2) = √216 ≈ 14.70 area units. No angle needed, no height; therefore Heron is the method of choice in the SSS case. Two checks pay off: first, the triangle inequality must hold (each side smaller than the sum of the others), otherwise the radicand is negative and the triangle does not exist. Second, you can verify by computing an angle with the law of cosines and cross-checking with ½·a·b·sin γ; here: cos γ = (25 + 36 − 49)/60 = 0.2, sin γ ≈ 0.98, A = 15·0.98 ≈ 14.7 ✓.

Yes, without modification. The sine is positive for all angles between 0° and 180°, so A = ½·a·b·sin γ yields a positive area even for obtuse γ. Indeed sin(180° − γ) = sin γ: a triangle with γ = 120° has the same sine value as one with 60°, namely ≈ 0.866. Example: a = 5, b = 7, γ = 120°: A = 17.5·0.866 ≈ 15.16 area units, exactly as large as for the 60° triangle with the same sides; the two triangles differ but have equal area. Intuitively: for an obtuse angle the height h = b·sin γ falls outside the triangle onto the extension of the base, but keeps its length. Only at γ = 0° or 180° does the triangle degenerate and A = 0.