Question

Keeping the origin constant axes are rotated at an angle ${30}^o$ in negative direction, if then new coordinates of the point are $(2,1)$ then its coordinate are with respect to old axis are

• A. $(\frac{2+\sqrt{3}+1}{2},\frac{2\sqrt{3}}{2})$
• B. $(\frac{2\sqrt{3}+1}{2},\frac{-2+\sqrt{3}}{2})$
• C. $(\frac{2+\sqrt{3}}{2},\frac{\sqrt{3}-2}{2})$
• D. $noneofthese$

Answer 1

The correct option is B $(\frac{2\sqrt{3}+1}{2},\frac{-2+\sqrt{3}}{2})$

Given $\theta ={30}^o$ in negative direction $(X,Y)=(2,1)$

New coordinates are given as:

$x=X\cos \theta -Y\sin \theta$

$y=X\sin \theta +Y\cos \theta$

Here $(x,y)$ are old coordinates

$\theta =-{30}^o$

So, substituting

$x=2\cos (-{30}^o)-1\sin (-{30}^o)$ $=2\frac{\sqrt{3}}{2}+\frac{1}{2}$

$x=\sqrt{3}+\frac{1}{2}=\frac{(2\sqrt{3}+1)}{2}$

$y=2\sin (-{30}^o)+1\cos (-{30}^o)$

$y=2\times \frac{-1}{2}+1\times \frac{\sqrt{3}}{2}$

$y=\frac{-2+\sqrt{3}}{2}$