Find the magnitude and direction of the resultant of two vectors $A$ and $B$ in terms of their magnitudes and angle $\theta$ between them.

Answer Let $OP$ and $O$ represent the two vectors
$A$ and $B$ making an angle $\theta$ . Then, using the parallelogram method of vector addition, $OS$ represents the resultant vector $R :$
$R = A + B$
$S N$ is normal to $OP$ and $P M$ is normal to $O S .$
From the geometry of the figure,
$O S^{2}=O N^{2}+S N^{2}$
but $\quad O N=O P+P N=A+B \cos \theta$
$S N=B \sin \theta$
$O S^{2}=(A+B \cos \theta)^{2}+(B \sin \theta)^{2}$
$R^{2}=A^{2}+B^{2}+2 A B \cos \theta$
$R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}$
In $\Delta$ $OSN$, $S N=O S \sin \alpha=R \sin \alpha,$ and
in $\Delta$ $PSN$, $\quad S N=P S \sin \theta=B \sin \theta$
Therefore, $\quad R \sin \alpha=B \sin \theta$
$\frac{R}{\sin \theta}=\frac{B}{\sin \alpha}$
Similarly,
$PM =A \sin \alpha=B \sin \beta$
$\frac{A}{\sin \beta}=\frac{B}{\sin \alpha}$
$\frac{R}{\sin \theta}=\frac{A}{\sin \beta}=\frac{B}{\sin \alpha}$
$\sin \alpha=\frac{B}{R} \sin \theta$
$\tan \alpha=\frac{S N}{O P+P N}=\frac{B \sin \theta}{A+B \cos \theta}$