Question

If the axes are rotated through an angle of ${30}^o$ in the clockwise direction, the point $(4,2\sqrt{3})$ in the new system is

• A. $(2,3)$
• B. $(2,\sqrt{3})$
• C. $(\sqrt{3},2)$
• D. $(\sqrt{3},5)$

Answer 1

The correct option is D $(\sqrt{3},5)$

We have, $x=4,y=2\sqrt{3}$ and $\theta =-{30}^o$
$\therefore X=x\cos \theta +y\sin \theta$
and $Y=-x\sin \theta +y\cos \theta$
$\Rightarrow X=4\cos {30}^o-2\sqrt{3}\sin {30}^o$
$=4\times \frac{\sqrt{3}}{2}-2\sqrt{3}\times \frac{1}{2}$
and $Y=4\sin {30}^o+2\sqrt{3}\cos {30}^o$
$=4\times \frac{1}{2}+2\sqrt{3}\times \frac{\sqrt{3}}{2}$
$\Rightarrow X=\sqrt{3}$ and $Y=5$
Hence, the given point in the new system is $(\sqrt{3},5)$.