Prove that in any $\Delta ABC,$ $\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$, where $a,b$ and $c$ are the magnitudes of the sides opposite to the vertices $A,B$ and $C$, respectively.
Here, components of $C$ are $c\cos A$ and $c\sin A$ is drawn.

Since, $\overrightarrow {CD} = b - c\cos A$
In $\Delta BDC$,

${a^2} = {(b - c\cos A)^2} + {(c\sin A)^2}$

$\Rightarrow$ ${a^2} = {b^2} + {c^2}{\cos ^2}A - 2bc\cos A + {c^2}{\sin ^2}A$

$\Rightarrow$ $2bc\cos A = {b^2} - {a^2} + {c^2}\left( {{{\cos }^2}A + {{\sin }^2}A} \right)$

$\therefore$ $\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$