Area of Triangle in Terms of Two Sides and Angle/Proof 1

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 $c$.

$\ds \map \Area {ABC}$

$=$

$\ds \frac 1 2 h c$

$\ds $

$=$

$\ds \frac 1 2 h \paren {p + q}$

$\ds $

$=$

$\ds \frac 1 2 a b \paren {\frac p a \frac h b + \frac h a \frac q b}$

$\ds $

$=$

$\ds \frac 1 2 a b \paren {\sin \alpha \cos \beta + \cos \alpha \sin \beta}$

Definition of Sine of Angle and Definition of Cosine of Angle

$\ds $

$=$

$\ds \frac 1 2 a b \, \map \sin {\alpha + \beta}$

$\ds $

$=$

$\ds \frac 1 2 a b \sin C$

$\blacksquare$