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 $c$.