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

$A$ 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