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(A) Given,magnitude of displacement vector $|\vec{d}| = 4$ and angle $	heta = 30^{\circ}$ with the $x$-axis.The rectangular components of a vector $\

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$2 \sqrt{3}, 2$

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(A) Given,magnitude of displacement vector $|\vec{d}| = 4$ and angle $	heta = 30^{\circ}$ with the $x$-axis.The rectangular components of a vector $\vec{d}$ are given by:$d_x = |\vec{d}| \cos 	heta$$d_y = |\vec{d}| \sin 	heta$Substituting the values:$d_x = 4 \cos 30^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3}$$d_y = 4 \sin 30^{\circ} = 4 	imes \frac{1}{2} = 2$Thus,the rectangular components are $2\sqrt{3}$ and $2$.

$A$ displacement vector of magnitude $4$ makes an angle $30^{\circ}$ with the $x$-axis. Its rectangular components in the $x-y$ plane are .........

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