Area of the Triangle

The familiar formula $\triangle=\tfrac12(\text{base})\times(\text{height})$ needs a height, and for an oblique triangle no height is directly given — so the height is first expressed using the sine of an angle.

Theorem 3.5 (Area formula). In $\triangle ABC$,

Proof. Drop the altitude from $A$ to $BC$, meeting it at $D$. In right triangle $ADC$, $\dfrac{AD}{AC}=\sin C\Rightarrow AD=b\sin C$. Taking $BC=a$ as the base and $AD=b\sin C$ as the height,

Dropping the altitude from $B$ or from $C$ instead gives the other two forms by the same argument. $\blacksquare$

Remarks.

1. In words: the area of a triangle is half the product of two sides and the sine of the angle included between them.
2. Neither the third side nor an altitude needs to be found by hand — the sine of the included angle supplies the height automatically. This is a genuine shortcut whenever an SAS data set is available.
3. It is itself a generalisation of the right-triangle area formula: when the included angle is $90^\circ$, $\sin 90^\circ=1$ and $\triangle=\tfrac12ab$, the ordinary "half base times height" formula.
4. Notice what is not needed: the third side plays no role at all in the formula, and there is no need to construct an altitude. Application — area of a circular segment. A segment of a circle is the region between a chord and the arc it cuts off. If a chord $AB$ subtends an angle $\theta$ (in radians) at the centre $O$ of a circle of radius $r$, then $\text{Area of segment}=\text{Area of sector }OAB-\text{Area of }\triangle OAB=\frac12r^2\theta-\frac12r^2\sin\theta=\frac12r^2(\theta-\sin\theta),$ …