Question

A line segment $AB$ of length $2$ units, where $A=(1,0)$ is rotated in anticlockwise direction through an angle ${15}^{\circ }$ about $A$. If initially $AB$ is making an angle ${45}^{\circ }$ with the positive direction of $X$- axis in anticlockwise direction and the distance between origin and the new position of $B$ is $k$units, then the value of $k^2$ is

Answer 1

Let $B^{\prime }$ is the new position of $B$,

Given that $A{B}^{\prime }$ makes an angle ${60}^{\circ }$ with the positive direction of $X$-axis.

$A=(1,0),AB=r=2,\theta ={60}^{\circ }$
Now,
$B^{\prime }=(x_1+r\cos \theta ,y_1+r\sin \theta )\Rightarrow B^{\prime }=(1+2\times \cos {60}^{\circ },0+2\times \sin {60}^{\circ })$
$\Rightarrow B^{\prime }=(1+2\times \frac{1}{2},0+2\times \frac{\sqrt{3}}{2})\Rightarrow B^{\prime }=(2,\sqrt{3})$

So, the distance of $B^{\prime }$ from origin
$=\sqrt{(2{)}^2+(\sqrt{3}{)}^2}=\sqrt{7}$ units
$\therefore k^2=7$