Find the resultant.


A vector $\vec{A}$ makes an angle of $20^{\circ}$ and $\vec{B}$ makes an angle of $110^{\circ}$ with the 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, $R^{2}=A^{2}+B^{2}+2 A B \cos \theta$

Now, $\mathrm{R}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta$

Now, $R^{2}=A^{2}+B^{2}+2 A B \cos \theta$

$=3^{2}+4^{2}+2.3 .4 \operatorname{Cos}(90)$


Or, $R=5$

Let $\phi$ is the angle between $R$ and $A$,

Then $\tan \phi=\frac{\frac{B \sin \theta}{A+B \cos \theta}}{}=\frac{4}{3}$, or $\phi=53^{\circ}$.

The resultant makes an angle of $(53+20)^{\circ}=73^{\circ}$ with the $x$ axis.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now