A vector $\vec{A}$ makes an angle of $20^{\circ}$ and $\vec{B}$ makes an angle of $110^{\circ}$ with the $\mathrm{X}$-axis. The magnitudes of these vectors are $3 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively. Find the resultant.
The angle between $\mathbf{A}$ and $\mathbf{B}$ from the x-axis are $20^{\circ}$ and $110^{\circ}$ respectively. Their magnitudes are 3 units and 4 units respectively.
Thus the angle between $\mathbf{A}$ and $\mathbf{B}$ is $=110-20=90^{\circ}$
Now, $R^{2}=A^{2}+B^{2}+2 A B \cos \theta$
$=3^{2}+4^{2}+2.3 .4 \operatorname{Cos}(90)$
$=5^{2}$
Or, $\mathrm{R}=5$
Let $\phi$ is the angle between $R$ and $A$,
Then $\tan \phi=\frac{\operatorname{Bsin} \theta}{A+B \cos \theta}=\frac{4}{3}$, or $\phi=53^{\circ}$.
The resultant makes an angle of $(53+20)^{\circ}=73^{\circ}$ with the $x$ axis.