Area of Triangle in Terms of Two Sides and Angle/Proof 2

Theorem

The area of a triangle $ABC$ is given by:

$\dfrac 1 2 a b \sin C$

where:

$a, b$ are two of the sides

$C$ is the angle of the vertex opposite its other side $c$.

Proof

By definition of sine:

$h = b \sin C$

From Area of Triangle in Terms of Side and Altitude:

$\map \Area {ABC} = \dfrac {a h} 2$

Substituting:

$\map \Area {ABC} = \dfrac {a b \sin C} 2$

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Area of the triangle