Angle (in rad) made by the vector $\sqrt{3}\hat{\mathrm{i}}+\hat{\mathrm{j}}$ with the x-axis is

Let $A=\sqrt{3}\hat{i}+\hat{j}$​ and $B=\hat{i}$ (unit vector along X-axis) 

A.B = $(\sqrt{3}\hat{i}+\hat{j}).\hat{j}$ = $\sqrt{3}+0$ = $\sqrt{3}$ 

Also, $\mid A\mid =\sqrt{{(\sqrt{3})}^2+{\left(1\right)}^2}=\sqrt{3+1}=2$

$\mid B\mid =1$

∵ ABcosθ = A.B where θ is the angle between vector A and X - axis. 

cosθ = A.B/AB ​=$\frac{\sqrt{3}}{2}$= cos(π/6)​

 θ = π/6​ rad