The figure shows an overhead view of a corridor with a plane mirror MN mounted at one end. A burglar B sneaks along the corridor directly towards the centre of the mirror. If d=3 m , how far from the mirror will the burglar be when the security guard S can first see him in the mirror?

Question

The figure shows an overhead view of a corridor with a plane mirror MN mounted at one end. A burglar B sneaks along the corridor directly towards the centre of the mirror. If d=3 m , how far from the mirror will the burglar be when the security guard S can first see him in the mirror? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-x9qxeou0babp.png" /> (A)  text3.0m (B)  text4.5m (C)  text6.0m (D)  text1.5m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-x9qxeou0isd3.png" /> tan θ=( mathrmX/(( mathrmMN)2)) (in the smaller triangle) tan θ =(3 m/3 m) (in the bigger triangle) ∴ (x/ text1.5)=(3/3) ∴ x= text1.5m

Q. The figure shows an overhead view of a corridor with a plane mirror $MN$ mounted at one end. A burglar $B$ sneaks along the corridor directly towards the centre of the mirror. If $d = 3 m$ , how far from the mirror will the burglar be when the security guard $S$ can first see him in the mirror?

$3.0 m$

$4.5 m$

$6.0 m$

$1.5 m$

$tan θ = \frac{X}{( \frac{MN}{2} )}$ (in the smaller triangle)
$t an θ = \frac{3 m}{3 m}$ (in the bigger triangle)
$∴ \frac{x}{1.5} = \frac{3}{3}$
$∴ x = 1.5 m$

Q. The figure shows an overhead view of a corridor with a plane mirror MN mounted at one end. A burglar B sneaks along the corridor directly towards the centre of the mirror. If d=3m , how far from the mirror will the burglar be when the security guard S can first see him in the mirror?

3.0m

4.5m

6.0m

1.5m