A particle starting from the origin (0, 0) mo | vedclass

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(D) Let $	heta$ be the angle that the path of the particle makes with the $x$-axis.From the given coordinates $(x, y) = (\sqrt{3}, 3)$,the slope of t

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) Let $	heta$ be the angle that the path of the particle makes with the $x$-axis.From the given coordinates $(x, y) = (\sqrt{3}, 3)$,the slope of the line is given by:$	an 	heta = \frac{y}{x} = \frac{3}{\sqrt{3}}$Simplifying the expression:$	an 	heta = \sqrt{3}$Taking the inverse tangent on both sides:$	heta = 	an^{-1}(\sqrt{3}) = 60^{\circ}$Thus,the path of the particle makes an angle of $60^{\circ}$ with the $x$-axis.

$A$ 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 an angle of $.......^{\circ}$ with the $x$-axis.

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