Let vector A and vector B be the two vectors of magnitude 10 unit each.

Question:

Let $\vec{A}$ and $\vec{B}$ be the two vectors of magnitude 10 unit each. If they are inclined to the $\mathrm{X}$-axis at angles $30^{\circ}$ and $60^{\circ}$ respectively, find the resultant.

Solution:

$A$ and $B$ are inclined at angles of 30 degrees and 60 degrees with respect to the $x$ axis

Angle between them $=(60-30)=90$ degrees

Given that $|\mathrm{A}|=|\mathrm{B}|=10$ units, we get

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

$=10^{2}+10^{2}+2.10 .10 \operatorname{Cos}(30)$

$R=20 \cos 15^{\circ}$

Or, $\mathrm{R}=19.3$ units

And $\tan \phi=\frac{B \sin \theta}{A+B \cos \theta}=\frac{1}{2+\sqrt{3}}$, or $\phi=15^{\circ}$.

Therefore, this resultant makes an angle of $(15+30)=45$ degrees with the $x$ axis

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