State parallelogram law of vectors. Derive an expression for the magnitude and direction of resultant vector?

If two vectors are represented in magnitude and direction by the two adjascent sides of a parallelogram drawn from a point then their resultant is represented in magnitude and direction by the diagonal passing through the same point.

Magnitude of Resultant :
Let two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ be represented in magnitude and direction by two adjacent sides $\overline{\mathrm{OA}}$ and $\overline{\mathrm{OB}}$ of parallelogram OACB. $\overrightarrow{\mathrm{R}}$ will be represented by the diagonal $\overline{\mathrm{OC}}$ of a parallelogram.
The line OA extended upto point N and draw a perpendicular from point ‘C’ on to the extended line. From parallelogram OACB
$\begin{array}{l}\mathrm{\angle }\mathrm{BOA}=\mathrm{\angle }\mathrm{CAN}=\mathrm{\theta }\\ \mathrm{OA}=\mathrm{P},\mathrm{OB}=\mathrm{Q}\text{ and }\mathrm{OC}=\mathrm{R}\text{ From }{\mathrm{\Delta }}^{\mathrm{le}}\mathrm{CAN}\end{array}$

$\begin{array}{l}\mathrm{Sin}\mathrm{\theta }=\frac{\mathrm{CN}}{\mathrm{AC}}\\ \Rightarrow \mathrm{CN}=\mathrm{Qsin}\mathrm{\theta }\\ \mathrm{Cos}\mathrm{\theta }=\frac{\mathrm{AN}}{\mathrm{AC}}\\ \Rightarrow \mathrm{AN}=\mathrm{Qcos}\mathrm{\theta }\end{array}$
From $∆$^leONC , applying phythagarous theorem
$\begin{array}{l}{\mathrm{OC}}^2={\mathrm{ON}}^2+{\mathrm{CN}}^2(\mathrm{ON}=\mathrm{OA}+\mathrm{AN})\\ {\mathrm{OC}}^2=(\mathrm{OA}+\mathrm{AN}{)}^2+{\mathrm{CN}}^2\\ {\mathrm{OC}}^2={\mathrm{OA}}^2+{\mathrm{AN}}^2+2\left(\mathrm{OA}\right)\left(\mathrm{AN}\right)+{\mathrm{CN}}^2\\ {\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2{\cos }^2\mathrm{\theta }+2\mathrm{PQcos}\mathrm{\theta }+{\mathrm{Q}}^2{\sin }^2\mathrm{\theta }\\ \therefore {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\\ {\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)+2\mathrm{PQcos}\mathrm{\theta }\\ \mathrm{R}=\sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQcos}\mathrm{\theta }}\end{array}$
Direction of Resultant:
From triangle ONC
$\begin{array}{l}\mathrm{Tan}\mathrm{\alpha }=\frac{\mathrm{CN}}{\mathrm{ON}}=\frac{\mathrm{CN}}{\mathrm{OA}+\mathrm{AN}}==\frac{\mathrm{Qsin}\mathrm{\theta }}{\mathrm{P}+\mathrm{Qcos}\mathrm{\theta }}\\ \mathrm{\alpha }={\tan }^{-1}\left(\frac{\mathrm{Qsin}\mathrm{\theta }}{\mathrm{P}+\mathrm{Qcos}\mathrm{\theta }}\right)\end{array}$