Triangle Area (Two Sides and Included Angle)

This is half base times height with the height supplied by trigonometry rather than by a tape. Take side $a$ as the base. Drop a perpendicular from the far end of side $b$ down to it; that perpendicular is the opposite leg of a right triangle whose hypotenuse is $b$ and whose angle is $C$, so its length is exactly $b\sin C$. Substitute into $A = \tfrac{1}{2}(\text{base})(\text{height})$ and you have $A = \tfrac{1}{2}ab\sin C$. Nothing new has been introduced — the formula's whole contribution is letting you skip a measurement that is awkward to take, because the height of a triangle usually hangs in mid-air.

Surveyors and land agents use it constantly. Two fence lines running 30 m and 42 m from a common corner, with 75° between them, enclose $\tfrac{1}{2}(30)(42)\sin 75° = 608$ m². The same arithmetic sizes a triangular section of roof, a gusset plate, or a wedge-shaped bay in a floor plan, and the solver runs it backwards to give the side a required area demands.

Two connections are worth carrying. Set $C = 90°$, where $\sin C = 1$, and the expression collapses to $\tfrac{1}{2}ab$ — the two sides are the legs of a right triangle. Since the sine can never exceed 1, that is also the largest triangle two given sides can make: for fixed $a$ and $b$, the area is greatest when they meet squarely, and falls away toward zero as the angle closes to 0° or opens toward 180°. The second connection is to vectors: $ab\sin C$ is the magnitude of the cross product of two edge vectors, so this formula is the cross product wearing a triangle costume, and it is what underlies the shoelace method for polygon areas.

The failure mode here is a quiet one, which makes it worse. $C$ must be the angle enclosed between $a$ and $b$, not simply an angle of the triangle you happen to know. Feed in a different angle and the arithmetic proceeds without complaint and returns a number that is wrong — there is no domain error to catch you, because nothing invalid happened. Check that the two sides you entered both radiate from the corner whose angle you entered. There is also a limit built into the geometry: the solver will not solve for $C$, because $\sin C = \sin(180° - C)$, so a 40° corner and a 140° corner between the same two sides give identical areas and the area cannot tell them apart. And if what you actually hold is three side lengths rather than two sides and an angle, this is the wrong page — Heron's formula is the one that takes three sides.