Area of Triangles (Proof)

Proof:
\[Area\;of\;\Delta ABC = \frac{1}{2}ch\]
Since
\[\sin \theta = \frac{b}{h}\]
\[h = b\sin \theta \]
Thus,
\[Area\;of\;\Delta ABC = \frac{1}{2}cb\sin \theta \]
where \(\theta \) is the angle between the sides \(b\) and \(c\).