If the axes are rotated by an angle of 30^{ci | vedclass

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(B) The transformation equations for rotating the axes by an angle $	heta$ are given by:$x = x' \cos 	heta - y' \sin 	heta$$y = x' \sin 	heta + y'

Vedclass · Accepted Answer

$\left( \frac{2\sqrt{3} - 1}{2}, \frac{2 + \sqrt{3}}{2} \right)$

Vedclass · Answer

(B) The transformation equations for rotating the axes by an angle $	heta$ are given by:$x = x' \cos 	heta - y' \sin 	heta$$y = x' \sin 	heta + y' \cos 	heta$Given $	heta = -30^{\circ}$ (negative direction),we have:$x = x' \cos(-30^{\circ}) - y' \sin(-30^{\circ}) = x' \frac{\sqrt{3}}{2} + y' \frac{1}{2}$$y = x' \sin(-30^{\circ}) + y' \cos(-30^{\circ}) = -x' \frac{1}{2} + y' \frac{\sqrt{3}}{2}$Given the original point $(x, y) = (2, 1)$,we solve for $(x', y')$:$2 = x' \frac{\sqrt{3}}{2} + y' \frac{1}{2} \implies 4 = x'\sqrt{3} + y' \quad (1)$$1 = -x' \frac{1}{2} + y' \frac{\sqrt{3}}{2} \implies 2 = -x' + y'\sqrt{3} \quad (2)$From $(2)$,$y'\sqrt{3} = 2 + x' \implies y' = \frac{2 + x'}{\sqrt{3}}$.Substituting into $(1)$:$4 = x'\sqrt{3} + \frac{2 + x'}{\sqrt{3}} \implies 4\sqrt{3} = 3x' + 2 + x' \implies 4x' = 4\sqrt{3} - 2 \implies x' = \frac{2\sqrt{3} - 1}{2}$.Substituting $x'$ into $(2)$:$y'\sqrt{3} = 2 + \frac{2\sqrt{3} - 1}{2} = \frac{4 + 2\sqrt{3} - 1}{2} = \frac{3 + 2\sqrt{3}}{2} \implies y' = \frac{\sqrt{3} + 2}{2}$.Thus,the new coordinates are $\left( \frac{2\sqrt{3} - 1}{2}, \frac{2 + \sqrt{3}}{2} ight)$.

If the axes are rotated by an angle of $30^{\circ}$ in the negative direction (clockwise) while keeping the origin fixed,what are the new coordinates of the point $(2, 1)$?

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