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,

$PM =A \sin \alpha=B \sin \beta$

$\frac{A}{\sin \beta}=\frac{B}{\sin \alpha}$

$\frac{R}{\sin \theta}=\frac{A}{\sin \beta}=\frac{B}{\sin \alpha}$

$\sin \alpha=\frac{B}{R} \sin \theta$

$\tan \alpha=\frac{S N}{O P+P N}=\frac{B \sin \theta}{A+B \cos \theta}$