When a plane mirror is placed horizontally on level ground at a distance of $60\text{m}$ from the foot of a tower, the top of the tower and its image in the mirror subtends an angle of ${90}^{\circ }$ at the corner of the mirror nearer to the foot of tower. The height of the tower is

1. $60\text{m}$
2. $90\text{m}$
3. $120\text{m}$
4. $30\text{m}$

The correct option is A $60\text{m}$
Let $h$ be height of tower $\left(AB\right)$


For a plane mirror,
Image size $\left(A^{\prime }B\right)$ = Object size $\left(AB\right)$,
and from figure, $MB$ is perpendicular on the line $A{A}^{\prime }$
$\Rightarrow$ Angle subtended by $A$ and $A^{\prime }$ at mirror's nearest point $\left(M\right)$ will be bisected by line or normal $MB$.
So, total subtended angle by $A{A}^{\prime }$ on point $M$$=\theta +\theta =2\theta$
Given that
$2\theta ={90}^{\circ }$ and $MB=60\text{m}$
$\therefore \theta ={45}^{\circ }$
In $\Delta AMB$
$\tan \theta =\tan {45}^{\circ }=\frac{AB}{MB}=\frac{AB}{60}$
$\Rightarrow AB=60\times \tan 45=60\text{m}$
$\therefore h=AB=60\text{m}$
Thus, option $\left(b\right)$ is the correct answer.