A line through the point $A(2,0)$ which makes an angle of ${30}^o$  with $+ve$ direction of $x-axis$  is rotated about A in clock wise direction through an angle ${15}^o$ . Then the equation of the straight line in the new position is

Equation of line which is passing through $(x_1,y_1)$ and angle of inclination $\theta$ with +ve x-axis in anticlockwise     
$\Rightarrow y-y_1=\tan \theta (x-x_1)$ $(y-0)=\tan {15}^o(x-2)$

$[\tan {15}^o=\tan ({45}^o-{30}^o)]=\frac{\tan 45-\tan {30}^o}{1+\tan {45}^o;\tan {30}^o}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

$\therefore (2-\sqrt{3})x-y-4+2\sqrt{3}=0$ .