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A particle starting from the origin $(0,0)$ moves in a straight line in the $(x, y)$ plane. Its coordinates at a later time are $(\sqrt{3}, 3)$. The path of the particle makes with the $x$-axis an angle of
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$60^{\circ}$
Draw the situation as shown. OA represents the path of the particle starting from origin $O(0,0)$ Draw a perpendicular from point $A$ to x-axis. Let path of the particle makes an angle $\theta$ with the x -axis, then
$\begin{aligned} \tan \theta & =\text { slope of line } O A \\ & =\frac{A B}{O B}=\frac{3}{\sqrt{3}}=\sqrt{3} \\ \text { or } \quad \theta & =60^{\circ}\end{aligned}$
$\begin{aligned} \tan \theta & =\text { slope of line } O A \\ & =\frac{A B}{O B}=\frac{3}{\sqrt{3}}=\sqrt{3} \\ \text { or } \quad \theta & =60^{\circ}\end{aligned}$
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